As a style preference, I’m going to drop all the arguments that are from the probability space.
Fix a probability space (\Omega, \mathcal F, P); we characterize the Brownian motion \{B_t\}_{t \geq 0} via the following properties:
Theorem: If a process satisfies the above, then there are \mu, \sigma^2 (respectively called the drift and the variance parameter, and \sigma is named the volatility) such that there exist B_t \sim N(\mu t, \sigma^2 t).
Def: A stochastic process \{B_t\}_{t \geq 0} is called a (one dimensional) Brownian motion (or Wiener process) starting from the origin with drift \mu and variance parameter \sigma^2 if B_t = 0 and the above three conditions are satisfied, with the imposition that B_t - B_s \sim N(\mu (t-s), \sigma^2 (t-s)).
Prop: If B_t is a Brownian motion with \mu = 0, \sigma^2 = 1 (a so-called standard Brownian motion), then Y_t = \sigma B_t + \mu t is a Brownian motion with parameters \mu, \sigma^2.
Proof: Obvious.
Pick a probability space (\Omega, \mathcal F, P) that is rich enough to support a countable collection of independent standard normal variables. If you are particular, the unit interval with Lesbegue measure is sufficient here.
The strategy is as follows: we define B_t for a countable dense set (in particular the dyadic rationals) of times using our precession of standard normals. then, we find some t \mapsto B_t that agrees on the dense set and is uniformly continuous and then extend by continuity.
Set D_n = \left\{ \frac{k}{2^n}, k = 0, 1, \dots, 2^n \right\} and D = \bigcup_{n=0}^\infty D_n; index our standard normals by \{N_{q}\}_{q \in D}, and set B_0 = 0, B_1 = N_1, and B_{1/2} = \frac{B_1 - B_0}{2} + \frac{1}{2}N_{1/2}. Just continue the same thing for every such dyadic, such that \{B_{1/2^n} - B_0, B_{2/2^n} - B_{1/2^n}, \dots, B_1- B_{(2^n-1) / 2^n} \} are all independent N\left(0, 2^{-n}\right).
Theorem: Almost surely, t \mapsto B_t, t \in D is uniformly continuous.
Proof: Set K_n = \sup \{ |B_s - B_t| \mid s,t \in D, |s - t| \leq 2^{-n}\}. We just need to show that K_n \to 0 as n \to \infty. In fact, something even stronger is true: for \alpha < \frac{1}{2}, \lim_{n \to \infty} 2^{\alpha n}K_n = 0. Morally speaking, just think that each Brownian increment is about its standard deviation, which is |t - s|^{1/2}.
Technically, however, we proceed as follows: set Y_n = \max\{B_{1/2^n} - B_0, B_{2/2^n} - B_{1/2^n}, \dots, B_1- B_{(2^n-1) / 2^n} \} and note that the union bound yields \begin{align*} P(Y_n \geq x) &\leq \sum_{j=1}^{2^n} P(|B_{j/2^n} - B_{(j-1)/2^n}| \geq x) \\ &= 2^nP(B_{1/2^n} \geq x) \\ &= 2^{n+1}P(B_1 \geq 2^{n/2} x). \end{align*} If we choose x_n such that \sum_{n=1}^\infty 2^{n+1}P(B_1 \geq 2^{n/2}x_n) < \infty, then by Borel-Cantelli shows that for sufficiently large n, Y_n \leq x_n almost surely. Do any reasonable bound you like on the tail of the normal distribution and take a sufficiently large x and call it a day. In particular, if you choose the easiest bound P(N \geq x) \leq Ce^{-x^2/2}, you can eventually get the bound:
Prop: \limsup_{n \to \infty} \frac{2^{n/2}}{\sqrt{n}}Y_n \leq \sqrt{2 \log 2}.
Proof: Look at the sum \sum_{n=1}^\infty P\left(Y_n > \sqrt{n} \cdot 2 ^{-n/2} \cdot \sqrt{2 \log 2 (1 + \epsilon)}\right) and apply Borel-Cantelli. In particular, we have \begin{align*} P(Y_n > x_n) &\leq \sum_{j=1^{2^n}} P(|B_{j/2^n} - B_{(j-1)/2^n}| > x_n) \\ &= 2^{n+1}P(B_{1/2^n} > x_n) \\ &= 2^{n+1}P\left(B_1 > \sqrt{n} \sqrt{2(\log 2 (1 + \epsilon)}\right) \\ &\leq C 2^n e^{-\frac{\sqrt{2\log 2 n(1+\epsilon)^2}}{2}} \\ &\leq C e^{-n\epsilon}. \end{align*}
Prop: Set K_n = \sup \{ |B_s - B_t| \mid s,t \in D, |s - t| \leq 2^{-n}\}; then, there is C such that with almost surely, \limsup_{n \to \infty} \frac{2^{n/2}}{\sqrt{n}}K_n \leq C.
Proof: It’s easy to see that K_n \leq 2 \sum_{j=n+1}^\infty Y_j (this is just the triangle inequality). Then for sufficiently large n, we get K_n \leq 2 \cdot 2 \sum_{j=n+1}^\infty 2^{-j/2}\sqrt{j} with full probability, and so \sup_{\substack{s, t \in D \\ s < t}} \frac{|B_t - B_s|}{\sqrt{(t-s)|\log((t-s)^{-1})|}} < \infty.
Now we may set B_t for t \in [0, 1] by B_t = \lim_{\substack{s \to t \\ s \in D}}B_s, and check that this is in fact a genuine Brownian motion, which is not bad; and of course this construction can extend to [0, \infty) easily as well.
Def: A function f: [0, 1] -> \mathbb R is called Hölder continuous of order \beta \geq 0 if there is some C < \infty such that for all s, t, |f(t) - f(s)| \leq C|t-s|^\beta. Futher, f is weakly Hölder continuous of order \beta if it is Hölder continuous of order \alpha for all \alpha < \beta. In both cases, we will say Hölder-\beta continuous for short.
Prop: Brownian motion paths are weakly Hölder-\frac{1}{2} continuous.
Proof: Omitted.
Theorem: The function t \mapsto B_t is nowhere differentiable almost surely.
Proof: Assume |f'(t)| < K; there exists \delta > 0 such that if |s - t| \leq \delta, |f(t) - f(s)| \leq 2K|s-t|; in particular there is an N such that for all n > N, |s - t| \leq n^{-1}, |r - t| \leq n^{-1}, |f(s) - f(r)| \leq 4Kn^{-1}. Then, set Z_{k, n} = \max \left\{ |B_{k/n} - B_{(k-1)/n}|, |B_{(k+1)/n} - B_{k/n}|, |B_{(k+2)/n} - B_{(k+1)/n}|\right\} and Z_n = \min \{ Z_{k, n} \mid k = 1, \dots, n \}.
If B is differentiable, then there is some M such that Z_n \leq Mn^{-1} for all n. Now set E_M to be the event that Z_n \leq Mn^{-1} for all sufficiently large; our theorem reduces to showing that P(E_M) = 0 for all M. In fact, we will show that \lim_{n \to \infty} P(Z_n \leq Mn^{-1}) = 0.
Consider the union bound \begin{align*} P(Z_n \leq Mn^{-1}) &\leq \sum_{j=1}^{n}P(Z(n,k) \leq Mn^{-1}) \\ &\leq nP\left( \max \left\{ |B_{1/n}|, |B_{2/n} - B_{1/n}|, |B_{3/n} - B_{2/n}|\right\} \right) \\ &\leq nP(|B_{1/n}| \leq Mn^{-1})^3 \\ &\leq nP(|B_1| \leq Mn^{-1/2})^3 \end{align*} and just do literally the stupidest estimate you can, e.g. just look at the density and say that the probability is bounded by 2CMn^{-1/2}, so that the above is sent to zero as n \to \infty.
Def: A filtration \{ \mathcal F \}_{t \geq 0} is an incerasing collection of sub \sigma-algebras. Further, we put \mathcal F_{\infty} = \bigcup_{t \geq 0} \mathcal F_t.
Def: A stochastic process \{X_t \}_{t \geq 0} is adapted to \{ \mathcal F_t \}_{t \geq 0} if for each t, X_t is \mathcal F_t-measurable.
Def: A process \{ B_t \}_{t \geq 0} is a standard Brownian motion start at 0 w.r.t. \{ \mathcal F_t \}_{t \geq 0} if
Def: A random variable \tau taking values in [0, \infty] is called a stopping time with respect to \{ \mathcal F_t \}_{t \geq 0} if for every t, the event \{\tau \leq t\} \in \mathcal F_t.
Examples: The following are all stopping times:
Def: If \tau is a stopping time, then \mathcal F_\tau is the \sigma-algebra corresponding to the collection of events A such that for each t, A \cap \{ \tau \leq t \} \in \mathcal F_t.
Def: For some stochastic process \{X_t\}_{t \geq 0} (or any other indexed set) with filtration \{F_t\}_{t \geq 0}, X_t has the Markov property if it saitisfies that E[f(X_t) \mid \mathcal F_s] = E[f(X_t) \mid \sigma(X_s)].
Def: In general, if X_t is a stochastic process and \tau is a stopping time, both adapted to \{ \mathcal F_t \}_{t \geq 0} with P(\tau < \infty) = 1, then X_t has the strong Markov property if X_{\tau + t} is independent of \mathcal F_{\tau}.
Prop: Suppose B_t is a Brownian motion and \tau is a stopping time, both with respect to \{ \mathcal F_t \}, and assume P(\tau < \infty) = 1. Set Y_t = B_{\tau + t} - B_{\tau}. Then Y_t is a Brownian motion independent of \mathcal F_t, e.g. Brownian motion has the strong Markov property and the new process is also a Brownian motion.
Proof: You proceed by doing successive approximations.
In particular, the following is clear for Brownian motion:
Prop: If \{B_t\}_{t \geq 0} is a Brownian motion, t a fixed time, and Y_s = B_{t+s} - B_t, then \{Y_s\}_{s \geq 0}, is a BM and independent of \mathcal F_t = \sigma\{B_s \mid s \leq t\}.
Theorem: Set B_t to a Brownian motion with drift zero; the reflection principle is that P\left(\max_{0 \leq s \leq t} B_s \geq a\right) = 2P(B_t \geq a). Set \tau_a = \min \{ s \mid B_s = a \}; we also have P(\tau_a \leq t) = 2P(B_t \geq a) or equivalently P(B_t \geq a \mid \tau_a \leq t) = \frac{1}{2}.
Prop: If 0 < r < s < \infty, q(r, s) = P(B_t = 0 \text{ for some } r \leq t \leq s) = 1 - \frac{2}{\pi} \arctan\left(\sqrt{\frac{r}{s - s}}\right).
Proof: First, I claim that q(r, s) = q(1, s/r) just by a change of variables, so we only need to compute q(t) = q(1, 1 + t). Set A = \{ B_s = 0 \text{ for some } 1 \leq s \leq 1 + t \}, so that \begin{align*} q(t) &= \frac{2}{\sqrt{2 \pi}}\int_0^\infty P(A \mid B_1 = x) e^{-x^2 / 2}dx. \end{align*} However, by the reflection principle, we have that \begin{align*} P(A \mid B_1 = x) &= P\left(\max_{0 \leq s \leq t} B_s \geq x \right) \\ &= P \left( \min_{0 \leq s \leq t} B_s \leq -x \right) \\ &= 2P(B_t \geq x) = 2P\left(B_1 \geq \frac{x}{\sqrt{t}}\right) \end{align*} upon which we can just compute the integral.
Corollary: One dimensional standard Brownian motion is (pointwise) recurrent (that is, the zero set of Brownian motion is unbounded).
Corollary: Since Y_t = t^{-1}B_{1/t} is a standard Brownian motion, this shows that for any \epsilon > 0, Z_\epsilon = \{ t \mid B_t = 0, 0 \leq t \leq \epsilon \} has more elements that just 0.
Def: A process \{ M_t \}_{t \geq 0} is a supermartingale (resp. submartingale) w.r.t. \{ \mathcal F_t \} if - E[|M_t|] < \infty, - M_t is \{ \mathcal F_t \} adapted, - and if E[M_t \mid \mathcal F_s] \leq M_s (resp. \geq M_s) almost surely for s \leq t. A process which is both a submartingale and supermartingale is just a martingale, it is continuous if M_t is a continuous function of t almost surely, and is square integrable (or simply L^2) if it has finite second moment for all t.
As an aside, I’m going to stop saying with probability 1 or almost surely because it’s annoying!
Prop: Brownian motion (without drift) is an L^2 continuous martingale.
Theorem (Kolmogorov Zero-One): Tail events of independent \sigma-algebras happen with probability 0 or 1.
Proof: Set \mathcal F_n = \sigma(X_1, \dots, X_n), and \mathcal T_n = \sigma(X_{n+1}, \dots), \mathcal F_\infty = \bigcup_{n} \mathcal F_n, and \mathcal T_\infty = \bigcap_n \mathcal T_n.
Then, one can see that if A \in \mathcal F_\infty and \epsilon > 0, there is n and A_n \in \mathcal F_n such that P(A_n \Delta A) < \epsilon; even better, there exists A_n independent of A \in \mathcal T_\infty such that P(A \Delta A_n) < \epsilon, and so we conclude that P(A) = P(A)P(A).
Theorem (Blumenthal Zero-One): Let B_t be a Brownian motion with the standard filtration, and set \mathcal F_{0+} = \bigcap_{\epsilon > 0} \mathcal F_{\epsilon}; then, if A \in \mathcal F_{0+}, either P(A) = 0 or 1.
Let B_t be a standard Brownian motion; a partition \Pi of [0, 1] is a sequence 0 = t_0 < t_1 < \dots < t_k = 1, and the mesh of the partition is just \| \Pi \| = \max_{i = 1, \dots, k} t_i - t_{i - 1}. Now take a sequence of partitions \Pi_n, e.g. 0 = t_{0, n} < \dots < t_{k_n, n}, and define the quantity Q(t, \Pi) = \sum_{t_{j} <= t} (B_{t_j} - B_{t_{j-1}})^2 and Q_n(t) = Q(t, \Pi_n) and Q_n = Q_n(1).
Theorem: If \| \Pi_n \| \to 0, then Q_n \to 1 in probability. Furthermore, if \sum_{n=1}^\infty \| \Pi_n \| < \infty, then almost surely \lim_{n \to \infty} Q_n = 1.
Proof: A simple computation gives us that E(Q_n) = \sum_{j=1}^{k_n}E[(B_{t_j} - B_{t_{j-1}})^2] = \sum_{j=1}^{k_n}(t_j - t_{j-1}) = 1 and \mathop{\mathrm{Var}}(Q_n) = \sum_{i=1}^{k_n} \mathop{\mathrm{Var}}((B_{t_j} - B_{t_{j-1}})^2) = \sum_{j=1}^{k_n} (t_j - t_{j-1})^2 \mathop{\mathrm{Var}}(B_1^2) = 2\sum_{j=1}^{k_n}(t_j - t_{j-1})^2. Then, \mathop{\mathrm{Var}}(Q_n) \leq \| \Pi_n \| 2 \sum_{j=1}^{k_n}(t_j - t_{j-1}) = 2 \| \Pi_n \| and P(|Q_n - 1| \geq \epsilon) \leq \frac{\mathop{\mathrm{Var}}(Q_n)}{\epsilon^2} \leq \frac{2 \| \Pi_n \|}{\epsilon^2}.
The latter half of the theorem follows from Borel-Cantelli.
Theorem: In general, if \sum_{n=1}^\infty \| \Pi_n \| < \infty, then almost surely we have Q_n(t) \to t.
Proof: With probability 1 this holds for rational t, but by construction Q_n(t) is monotone, so it holds everywhere.
Def: In general, if X_t is a process, then its quadratic variation is \left\langle X\right\rangle_t = \lim_{n \to \infty} \sum_{t_{j,n} \leq t} (X_{t_{j, n}} - X_{t_{j-1, n}})^2 (sort of, since sometimes this depends on the partition).
Def: Alternatively, if M_t is a continuous L^2-martingale, its quadratic variation is the unique predictable process \left\langle M \right\rangle_t that makes M_t^2 - \left\langle M \right\rangle_t a martingale (in particular, this exists by Doob decomposition).
We showed above that \left\langle B_t \right\rangle_t = t.
Prop: If B_t is a standard Brownian motion and Y_t = \mu t + \sigma B_t, then \left\langle Y \right\rangle_t = \sigma^2 t.
Proof: Just check directly.
Lemma (Relaxed Borel-Cantelli): Let A_1, A_2, \dots be a sequence of events, and set \mathcal F_n = \sigma(A_1, \dots, A_n); if there is q_n with \sum_{n=1}^\infty q_n = \infty such that P(A_n \mid \mathcal F_{n-1}) \geq q_n, then A_n happens infinitely often almost surely.
Proof: Same as usual, but just with a little more caution.
Recall that the second Borel-Cantelli lemma tells us that if they are independent and \sum_{n=1}^\infty P(A_n) = \infty, then A_n occurs infinitely often with probability 1; this lemma is stronger than that.
Theorem: If B_t is a standard Brownian motion, then \limsup_{t \to \infty} \frac{B_t}{\sqrt{2 t \log \log t}} = 1.
Corollary: By symmetry, \liminf_{t \to \infty} \frac{B_t}{\sqrt{2 t \log \log t}} = -1.
Proof: Let \mathcal T_t = \sigma\{ B_{s + t} - B_t \mid s \geq 0\}, and \mathcal T_\infty = \bigcap_t \mathcal T_t; one can adapt the arguments from the Kolmogorov 0-1 law to show that everything in \mathcal T_\infty happens with probability 0 or 1. Then, A_\epsilon = \left\{ \omega \mid \limsup_{t \to \infty} \frac{B_t}{\sqrt{2 t (1 + \epsilon) \log \log t}} \leq 1\right\} is a tail event (e.g. in \mathcal T_\infty) and thus P(A_\epsilon) = 0 or 1. In fact, if \epsilon < 0 then it’s 0, and if \epsilon > 0 then it’s 1. In fact, by this 0-1 law and symmetry one may see P \left( \limsup_{t \to \infty} \frac{|B_t|}{\sqrt{2 t (1 + \epsilon) \log \log t}} \leq 1 \right) = P \left(\limsup_{t \to \infty} \frac{B_t}{\sqrt{2 t (1 + \epsilon) \log \log t}} \leq 1\right) as well.
First take \epsilon > 0 and take some \rho > 1 to be specified later. Then, let V_n = \left\{ |B_{\rho^n}| \geq \sqrt{2 \rho^n(1-\epsilon) \log \log \rho^n} \right\}; we want to to show that V_n occurs infinitely often. I claim that P(V_{n+1} \mid V_1, \dots, V_n) \geq P \left( B_{\rho^{n+1}} - B_{\rho^n} \geq \sqrt{2 \rho^{n+1}(1-\epsilon) \log \log \rho^n} \right). To see this, compute \begin{align*} &P \left( \frac{B_{\rho^{n+1}} - B_{\rho^n}}{\sqrt{\rho^{n+1} - \rho^n}} \geq \frac{\sqrt{2\rho^{n+1}(1-\epsilon)\log \log \rho^n}}{\sqrt{\rho^n (\rho - 1)}} \right) \\ &= P \left( \frac{B_{\rho^{n+1}} - B_{\rho^n}}{\sqrt{\rho^{n+1} - \rho^n}} \geq \sqrt{\frac{2 \rho}{\rho - 1}(1-\epsilon)(\log n + \log \log \rho)} \right) \\ \end{align*} and choose \rho large enough so that \frac{2 \rho}{\rho - 1}(1-\epsilon) < 1, and use the estimate P(B_1 > x) \sim \exp(-x^2 / 2) and conclude by the earlier lemma. The other direction for \epsilon < 0 is similar (in fact, easier since we may conclude from the first Borel-Cantelli lemma).
Def: Set B_t a standard Brownian motion, Z = \{ t \mid B_t = 0 \}, and Z_t = Z \cap [0, t]. Then, t \in Z is right-isolated if t \in Z, and \exists \epsilon > 0 such that (t, t + \epsilon) \cap Z = \emptyset; similar for left-isolated. A point which is both left and right isolated is just isolated.
With probability 1, 0 is not right-isolated; further, Z_1 is homeomorphic to the Cantor set.
With probability 1, the sets of left and right isolated points are countable, and there are no isolated points.
We can show that if q \in \mathbb{Q}_{\geq 0}, P(B_q = 0) = 0, and \tau_q = \min \{ t \geq q \mid B_t = 0\} is a stopping time; further, the left-isolated points are just \{ \tau_q \mid q \in \mathbb{Q}_{\geq 0}\}. And by strong Markov property, no \tau_q is a right-isolated point, so there are no isolated points.
Set \sigma_q = \max_t \{ t < q \mid B_t = 0 \} (this is well-defined, but not a stopping time). Then associate every q to an interval (\sigma_q, \tau_q); then Z = [0, \infty) \setminus \bigcup_q (\sigma_q, \tau_q) and has Lebesgue measure 0. To see this, we just interchange integrals: E[ \lambda(Z_1) ] = E\left[ \int_0^1 1_{\{B_s = 0\}} ds \right] = \int_0^1 P\{ B_s = 0 \}ds = 0.
Look at Z \cap [1, 2]; now cover [1, 2] by intervals of length n^{-1}; put the number of these intervals that intersect Z as X_n. Then, E[X_n] = \sum_{j=1}^n P\left(Z \cap \left[1 + \frac{j-1}{n}, 1 + \frac{j}{n} \right]\neq \emptyset \right) \sim Cn^{1/2}.
The box or Minkowski dimension of a set is given by the exponent above (sort of).
Def: The Hausdorff dimension comes from the Hausdorff measure: for \epsilon > 0 and \alpha, set \mathcal H^\alpha_\epsilon = \left\{ \inf \sum_{j=1}^\infty (\operatorname{diam} U_j)^\alpha \right\} where the infimum is over all coverings \bigcup_j=1^\infty U_j of V with each \operatorname{diam} (U_j) < \epsilon. Then the Hausdorff measure is given by \mathcal H^\alpha(V) = \lim_{\epsilon \to 0} \mathcal H_{\epsilon}^\alpha (V). Then, there is some number D such that \mathcal H^{\alpha}(V) = \begin{cases} \infty & \alpha < D \\ 0 & \alpha > D \end{cases} and we call this D the Hausdorff dimension. In general, the Hausdorff dimension is at most the Minkowski dimension, but in this case we actually do have equality.
The local time is the amount of time the Brownian motion spends at 0 by a certain time. It is sort of like the Cantor measure, insofar as if s < t, then L_t - L_s > 0 \iff (s, t) \cap Z \neq \emptyset.
Def: We define the local time of Brownian motion at x \in \mathbb{R}, L_t, by first setting L_{t, \epsilon}(x) = \frac{1}{2\epsilon} \int_0^t 1_{\{|B_s - x| \leq \epsilon\}} ds and then letting L_t(x) = \lim_{\epsilon \to 0} L_{t, \epsilon}. We abbreviate L_t = L_t(0).
We can compute the expectation \begin{align*} E[L_{t, \epsilon}] &= \frac{1}{2\epsilon} \int_0^t P(|B_s| \leq \epsilon)ds \\ &\sim \frac{1}{2\epsilon} \int_0^t 2e \frac{1}{2 \pi s}ds \\ &= \sqrt{\frac{2}{\pi}}t^{1/2} \end{align*}.
Theorem: With probability 1, L_t exists for all t, and this holds in L^2 as well.
There are more facts: L_t is continuous in t and non-decreasing, L_t - L_s > 0 \iff (s, t) \cap Z \neq \emptyset, and L_t is weakly \frac{1}{2}-Hölder continuous.
Theorem (Scaling Rule): L_t has the same distribution as t^{1/2} L_1. Further, M_t = \max_{ 0 \leq s \leq t}B_s has the same distribution as L_t.
Def: If B_t^1, \dots, B_t^d are independent standard Brownian motions, then B_t = (B_t^1, \dots, B_t^d) is a standard Brownian motion in \mathbb{R}^d.
Prop: B_0 = \mathbf 0, has independent increments, if s < t, B_t - B_s \sim N(0, (t-s)I), and B_t has continuous paths.
In particular, if we have the above with B_t - B_s \sim N(\mu, \Gamma), then B_t is a Brownian motion with dift \mu \in \mathbb{R}^d and covariance matrix \Gamma. Further, if AA^T = \Gamma, then we may write B_t = AY_t + t \mu for a standard Brownian motion Y_t.
Consider the open annulus with inner radius r and outer radius R, e.g. D(r, R) = \{ y \in \mathbb{R}^d \mid r < |y| < R \}. Start a Brownian motion at x, and let \tau = \tau(r, R) be the first time with |B_t| = r or |B_t| = R. What is the probability that |B_\tau| = R?
In one dimension, this is easy: stop the Brownian motion at \tau and look at what happens at infinity.
In higher dimensions, we need a quick detour.
For this section, a domain is an open connected subset of \mathbb{R}^d.
Def: For a domain D, we say f: D \to \mathbb{R} is harmonic if it is continuous (or merely locally integrable) and satisfies the mean value property f(x) = MV(f, x, \epsilon) = \int_{|x - y| = \epsilon} f(y) ds where s is the surface measure, normalized so that \int_{|x - y| = \epsilon}ds = 1.
In a probabilistic vein, if B_t is a standard d-dimensional Brownian motion starting at x, and \tau = \min \{ t \mid |B_t - x| = \epsilon \}, since B_t is rotation invariant, the above is just MV(f, x, \epsilon) = E[f(B_\tau)].
Def: The Laplacian of f is \nabla f = \sum_{j=1}^d \frac{\partial ^2 f }{\partial x_j^2}.
Prop: If f is C^2 in D, then \frac{1}{2d}\nabla f = \lim_{\epsilon \to 0} \frac{MV(f, x, \epsilon) - f(x)}{\epsilon^2}.
Proof: Look at the Taylor expansion.
Theorem: f is harmonic in D if and only if it is in C^2 and \nabla f = 0 everywhere.
Let \tau = \tau_{r, R} = \min \{ |B_t| = r \text{ or } R \} = \min \{ t \mid B_t \in \partial D \}. We will let a superscript x denote that B_0 = x, and let \varphi(x) = P^x(|B_\tau| = R). By rotational invariance, \varphi(x) = \varphi(|x|), and it is continuous at the boundary; we also clearly have \varphi(x) = 0 if |x| = r and \varphi(x) = 1 if |x| = R. The strong Markov property furnishes that \varphi is harmonic. Set \tau_\epsilon = \min \{ t \mid |B_t - x| = \epsilon\}; then P^X(|B_\tau| = R \mid \mathcal F_{\tau_\epsilon}) = \varphi(B_{\tau_\epsilon}) and P^X(|B_\tau| = R) = E^x[P(|B_\tau| = R \mid \mathcal F_{\tau_\epsilon})] = E^x[\varphi(B_{\tau_\epsilon})] = MV(\varphi, x, \epsilon). This list of properties gives a unique function (solve an ODE in polar coordinates), and is given by \varphi(x) = \frac{|x|^{2-d} - r^{2-d}}{R^{2-d} - |r|^{2-d}} for d \neq 2 and \varphi(x) = \frac{\log|x| - \log r}{\log R - \log r} for d = 2.
Let B_t be a standard Brownian motion in \mathbb{R}^d.
Theorem: With probability 1, Brownian motion is transient in dimensions at least 3; that is, \lim_{t \to \infty} |B_t| = \infty.
Proof: If d \geq 3, and we start at x with |x| > r, if we set T_r = \min \{ t \mid |B_t| = r \}, P^x(T_r < \infty) = \lim_{R \to \infty} P^X(|B_{\tau_{r, R}}| = r) = \left( \frac{r}{|x|} \right)^{d-2} < 1
Theorem: With probability 1, Brownian motion is neighborhood recurrent; that is, \forall z \in \mathbb{R}^2, \epsilon > 0, Brownian motion visits the disk of radius \epsilon about z infinitely often. However, it is not point recurrent, e.g. it hits z \neq 0 with probability 0.
Proof: If d = 2, then P^x(T_r < \infty) = \lim_{R \to \infty} P(|B_{\tau_{r, R}}| = r) = 1 But if T = \min\{ t \mid B_t = 0 \}, P^x(T < \infty) \leq \lim_{R \to \infty} \lim_{r \to 0} P^x(|B_{\tau_{r, R}}| = r) = 0
A fun fact is that if d \geq 2, then \{ B_t, t \geq 0 \} has Hausdorff dimension 2 but also zero Hausdorff-2 measure.
Take a bounded domain D \subset \mathbb{R}^d, and a continuous function F: \partial D \to \mathbb{R}; the Dirichlet problem is to find the unique continuous f: \overline D \to \mathbb{R} that agrees with F on \partial D and is harmonic on D.
In fact, uniqueness follows from the maximum principle, which says that the maximum of f is attained on the boundary (think about the mean value principle). Then subtract two solutions and see that it is 0.
Let T = \min\{t \mid B_t \in \partial D \}, and f(x) = E^x[F(B_T)]. This satisfies the mean value principle, and in fact continuous (which is kinda hard) but it is obviously locally integrable, so f here is harmonic.
In general, such a harmonic function is not necessarily continuous on the boundary: take the example of the punctured unit disk, with F(x) = 1 for |x| = 1 and F(0) = 0. And so everything is fine except at the origin, since f(x) = P^x(|B_T| = 1) = 1.
Def: If x \in \partial D, let \sigma = \inf \{ t > 0 \mid B_t \in \partial D\}; x is regular if P^x(\sigma = 0) = 1.
Prop: f as defined above is continuous at every regular boundary point.
Therefore the Dirichlet problem has a solution for every continuous F if and only if every point on \partial D is regular.