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The union bound gives $\mathbb{P}({\bigcup }_{n\ge m}{A}_n)\le {\sum }_{n\ge m}\mathbb{P}(A_n)\to 0$. Taking the decreasing intersection proves the first Borel-Cantelli lemma.

If $X_n\to X$ almost surely, then $1_{\{|X_n-X|>\epsilon \}}\to 0$ almost surely; dominated convergence theorem gives convergence of its expectation, hence convergence in probability.

If convergence in probability holds, any subsequence has a further one with $\mathbb{P}(|X_{n(k_r)}-X|>2^{-r})<2^{-r}$. Borel-Cantelli then gives almost-sure convergence. Conversely, failure in probability supplies a subsequence with probabilities bounded below by some positive constant, and no further subsequence can converge almost surely because that would imply convergence in probability.

After replacing $X_j$ by $X_j-\mu$, the maximal inequality for independent averages givesThis follows from the Doob Lp maximal inequality by dyadically grouping the partial sums. The strong law makesalmost surely. The displayed maximal function is in $L^p$, so dominated convergence applied to its $p$th power proves convergence in $L^p$.

Use the conventionThe Riemann-Lebesgue lemma states that if $f\in L^1({\mathbb{R}}^n)$, then $\hat{f}$ is continuous andContinuity follows from the dominated convergence theorem, since $e^{-ix\cdot {\xi }_j}\to e^{-ix\cdot \xi }$ and the integrands are dominated by $|f(x)|$.

For the decay, first take $g\in C_c^1({\mathbb{R}}^n)$. Choose a coordinate $j$ for which $|{\xi }_j|\ge |\xi |/\sqrt{n}$. Integration by parts givesand henceSince $C_c^1({\mathbb{R}}^n)$ is dense in $L^1({\mathbb{R}}^n)$, choose $g$ with $\Vert f-g{\Vert }_1<\epsilon$. The elementary Fourier bound givesso the decay for $g$ implies the decay for $f$.

The Parseval identity says that for $f,g\in L^2({\mathbb{R}}^n)$, with their Fourier transforms defined by the Plancherel theorem,In particular,

For the given radial function, as $|x|\to 0$,while as $|x|\to \infty$,Using polar coordinates, local integrability of $|f{|}^q$ is therefore determined byand integrability at infinity byThe assumptions $2a>-n$ and $b>n$ imply both $f\in L^2({\mathbb{R}}^n)$ and $f\in L^1({\mathbb{R}}^n)$. The Riemann--Lebesgue lemma and the direct estimate $|\hat{f}|\le \Vert f{\Vert }_1$ give $\hat{f}\in L^{\infty }$, while Parseval gives $\hat{f}\in L^2$. Finally, for every $2\le p<\infty$,Thus

Because the centered normal distribution density $g_t$ is nonnegative and has integral one, Fubini's theorem and translation invariance giveThus $f\ast g_t\in L^1$.

Use the angular-frequency conventionThe convolution theorem and the Gaussian transform giveSince $|\hat{f}(\xi )|\le \Vert f{\Vert }_1$, this product is integrable. Moreover, absolute integrability justifies exchanging the integrals:This proves the Fourier inversion theorem for the convolution directly.

Now additionally suppose $\hat{f}\in L^1$. For every $x$, the dominated convergence theorem givesFourier inversion identifies the right-hand side with $f(x)$ for Lebesgue almost everywhere $x$. Therefore the Gaussian approximate identity satisfies

Mazur theorem says that the weak closure and norm closure of a convex subset of a real or complex normed space coincide. The norm closure is contained in the weak closure because every norm-continuous linear functional is norm-continuous. Conversely, if $x$ is outside the norm closure of a convex set $C$, the Hahn-Banach separation theorem gives $f\in X^{\ast }$ and a real number $\alpha$ such thatThis weakly open separation shows that $x$ is outside the weak closure.

By definition,For each $n$, regard $x_n$ as the bounded functional $J{x}_n:X^{\ast }\to \mathbb{F}$ given by $J{x}_n(f)=f(x_n)$. Pointwise convergence makes the family $(J{x}_n)$ pointwise bounded on the Banach space $X^{\ast }$. The Uniform boundedness principle gives

Suppose first that $x_n\rightharpoonup 0$. Every subsequence indexed by an infinite set $M$ also converges weakly to zero. Thus zero belongs to the weak closure of the convex hull of $\{x_n:n\in M\}$, and Mazur's theorem puts it in its norm closure. This gives the required finite convex combination of norm below any prescribed $\epsilon$.

Conversely, if weak convergence fails, some $f\in X^{\ast }$ and an infinite subsequence satisfy either $\mathrm{Re}f(x_n)\ge \epsilon$ throughout or $\mathrm{Re}f(x_n)\le -\epsilon$ throughout. Every convex combination from that subsequence then has norm at least $\epsilon /\Vert f\Vert$, contradicting the stated property.

Now let the $x_n\in {\ell }^p$, $1<p<\infty$, have pairwise disjoint supports and satisfy $\Vert x_n{\Vert }_p\le C$. For $N$ distinct terms,The convex-combination criterion therefore proves $x_n\rightharpoonup 0$. This is weak convergence of bounded disjointly supported sequences in lp.

The final implication for a commutative unital C*-algebra is true. By the Commutative Gelfand--Naimark theorem, $A\cong C(K)$ and its characters are the point evaluations. The hypotheses say that the uniformly bounded functions $a_n$ converge pointwise to zero. Every functional on $C(K)$ is integration against a finite regular measure by the Riesz-Markov-Kakutani representation theorem; the dominated convergence theorem givesHence $a_n\rightharpoonup 0$.

The Martingale convergence theorem gives an almost-sure limit $M_{\infty }$, because $|M_n|\le 1$. The dominated convergence theorem also gives $M_n\to M_{\infty }$ in $L^1$.

To identify the limit, take $A\in {\bigcup }_n{\mathcal{F}}_n$. For some $N$, $A\in {\mathcal{F}}_N$, and for every $n\ge N$,Passing to the $L^1$ limit preserves this equality. The sets for which $\mathbb{E}[M_{\infty }{1}_A]=\mathbb{E}[Z{1}_A]$ form a monotone class containing the algebra ${\bigcup }_n{\mathcal{F}}_n$, so the equality holds throughout ${\mathcal{F}}_{\infty }=\sigma ({\bigcup }_n{\mathcal{F}}_n)$. Since $M_{\infty }$ is ${\mathcal{F}}_{\infty }$-measurable, it is $\mathbb{E}[Z\mid {\mathcal{F}}_{\infty }]$. This proves the conditional-expectation convergence along a filtration both almost surely and in $L^1$.

Define $Z_n={\sup }_{m\ge n}|X_m-X|$. Then $0\le Z_n\le 2$ and $Z_n\downarrow 0$ almost surely. The dominated convergence theorem gives $\mathbb{E}{Z}_n\to 0$.

Set $W_n=\mathbb{E}[Z_n\mid {\mathcal{F}}_n]$. Since $Z_{n+1}\le Z_n$, the tower property of conditional expectation givesso $(W_n)$ is a nonnegative supermartingale. The almost sure supermartingale convergence theorem gives $W_n\to W_{\infty }$ almost surely, and Fatou lemma yields $\mathbb{E}{W}_{\infty }\le {\mathrm{liminf}}_n\mathbb{E}{W}_n=0$. Hence $W_n\to 0$ almost surely; because $\mathbb{E}{W}_n=\mathbb{E}{Z}_n\to 0$, convergence also holds in $L^1$.

Finally,The second term tends to zero almost surely and in $L^1$ by part b. The first does so by the preceding argument, proving the moving-variable conditional-expectation convergence.

The almost sure supermartingale convergence theorem says that a supermartingale whose negative parts have uniformly bounded expectations converges almost surely to a finite integrable limit. In particular, every nonnegative supermartingale converges almost surely.

Here the process is uniformly bounded, say $0\le X_n\le C$. Let $X_{\infty }={\lim }_n{X}_n$, whose existence follows from the theorem. The dominated convergence theorem then givesThe limit on the right also exists directly because the expectations of a supermartingale form a decreasing sequence.

Set $g_{ϵ}(x)=x/\sqrt{{ϵ}^2+x^2}$ and $g(x)=\mathrm{sgn}(x)$, with $g(0)=0$. For every $s>0$, the normal distribution of $B_s$ has no atom at zero, so $g_{ϵ}(B_s)\to g(B_s)$ almost surely. Since $|g_{ϵ}-g|\le 2$, the dominated convergence theorem and Tonelli theorem giveThe Itô isometry followed by the Doob L2 maximal inequality now yieldsThus the stochastic integrals converge ucp to ${\int }_0^t\mathrm{sgn}(B_s)d{B}_s$.

For Brownian motion started at $x\in D$, the Itô formula and $\Delta u=0$ show thatis a local martingale. Since $D$ is bounded and $u$ is continuous on the compact set $\overline{D}$, the stopped process is bounded and hence a true martingale. Brownian motion exits every bounded domain almost surely, so $t\wedge \tau \to \tau$. The dominated convergence theorem, continuity at the boundary, and $u=f$ on $\partial D$ give the Brownian representation of the Dirichlet problem

Let $\sigma$ be the first time Brownian motion started outside the unit half-disc reaches its semicircular boundary. A path that reaches $A$ must first cross that semicircle. The Strong Markov property at $\sigma$ givesTake $z=iy$, multiply by $y$, and let $y\to \infty$. The Brownian representation of half-plane capacity identifies the left side with $\mathrm{hcap}(A)$, while part b givesSince the exit height is between zero and one, the dominated convergence theorem applies and yields

Let $T$ be a measure-preserving transformation of $(X,\mathcal{F},\mu )$ and let $f\in L^1(\mu )$. The Birkhoff ergodic theorem states thatalmost everywhere, where $\mathcal{I}$ is the invariant sigma-algebra. The limit is $T$-invariant and

Assume now that $\mu (X)=1$. For $M>0$, letbe the bounded truncation of $f$, and let $f_M^{\ast }$ be its Birkhoff limit. Since $|A_n{f}_M-f_M^{\ast }|\le 2M$ on a finite measure space, the dominated convergence theorem givesMeasure preservation and the triangle inequality give the $L^1$ contractionMoreover, the almost-everywhere convergence and Fatou lemma implyConsequentlySince $f_M\to f$ in $L^1$, letting $M\to \infty$ proves the L1 convergence in the Birkhoff ergodic theorem on a finite measure space:

For $0<\pi <1$, let ${\delta }_{\pi }$ be the Bayes classifier with class-zero prior $\pi$, and define its two class-conditional error probabilities byWriting the likelihood ratio as $L=f_1/f_0$, the rule chooses class one whenThe Gaussian likelihood-ratio level sets have probability zero, so dominated convergence shows that $e_0$ and $e_1$ depend continuously on $\pi$. As $\pi \downarrow 0$, the threshold tends to zero and the classifier chooses class one almost surely, givingAs $\pi \uparrow 1$, it chooses class zero almost surely, giving the opposite limit $(0,1)$. The intermediate value theorem therefore provides ${\pi }^{\ast }\in (0,1)$ such that

The rule ${\delta }_{{\pi }^{\ast }}$ is an equalizer rule with worst-case risk $r$. For any classifier $\delta$,because ${\delta }_{{\pi }^{\ast }}$ minimizes the integrated risk for its prior. Hence ${\delta }_{{\pi }^{\ast }}$ is a minimax decision rule, as summarized by the minimax Gaussian Bayes classifier from equal class errors.

The prior is indeed least favorable. Its Bayes risk is $r$, while for any other prior $q$ the Bayes risk is at most the integrated risk of the same equalizer rule ${\delta }_{{\pi }^{\ast }}$, namelyThus no prior has larger Bayes risk.

Regard the space of continuous functions vanishing at infinity $C_0({\mathbb{R}}^n)$ as a subspace of $L^{\infty }({\mathbb{R}}^n)$. Point evaluation at the origin,is a bounded linear functional of norm one on this subspace because a continuous function's supremum and essential supremum agree. By the Hahn-Banach theorem, ${\ell }_0$ extends to a bounded linear functional

Suppose that some $g\in L^1({\mathbb{R}}^n)$ represented this extension:Choose $\phi \in C_c({\mathbb{R}}^n)$ with $0\le \phi \le 1$, $\phi (0)=1$, and support in the unit ball, and set ${\phi }_k(x)=\phi (kx)$. Then $\ell ({\phi }_k)=1$ for every $k$. On the other hand, ${\phi }_k(x)\to 0$ for almost every $x$ and $|{\phi }_{k}g|\le |g|$. The dominated convergence theorem givesa contradiction. Thus $\ell$ is a singular functional on L infinity and has no $L^1$ density.

Multiply $\Delta u+qu\le 0$ by ${\zeta }^2/u_{\epsilon }$ and integrate. Since $\zeta$ is compactly supported, integration by parts and Young's inequality giveAs $\epsilon \downarrow 0$, $u/u_{\epsilon }$ converges to the indicator of $\{u>0\}$. The dominated convergence theorem therefore gives

Fix $t>0$ and define $w(s,y)=u(t-s,|y|)$ for $0\le s\le t$. The assumed smooth extension and the Neumann boundary condition at zero make $w$ a $C^{1,2}$ function. The heat equation givesThe Itô formula therefore makes $w(s,B_s)$ a local martingale. Stop first when $|B|$ leaves a large compact interval. The exponential growth bound and the finite exponential moments of the maximum of Brownian motion on $[0,t]$ give uniform integrability, so localization and the dominated convergence theorem yieldThis is the Feynman-Kac formula for the Neumann heat problem.

Let $p_t(z)=(2\pi t{)}^{-1/2}{e}^{-z^2/(2t)}$ be the heat kernel for $u_t=\frac{1}{2}{u}_{xx}$. The symmetry of the Gaussian distribution gives the method-of-images formulaThis is the Neumann heat kernel on a half-line. Differentiation under the integral sign shows that $v\in C^{1,2}$ for $t,x>0$ and that $v_t=\frac{1}{2}{v}_{xx}$. At $x=0$, the two differentiated kernel terms cancel, so $v_x(t,0)=0$. The Gaussian approximate identity gives $v(t,x)\to f(x)$ as $t\downarrow 0$, while the dominated convergence theorem gives continuity up to $x=0$. Finally $|v(t,x)|\le \Vert f{\Vert }_{\infty }$, which is stronger than the required exponential bound. Thus $v$ satisfies every condition in the displayed boundary-value problem.

For $f\in L^1(\mathbb{R})$, its regular tempered distribution acts by integration, and the distributional transform agrees with the functionThe triangle inequality gives the uniform boundIf ${\lambda }_j\to \lambda$, then $f(x)e^{-i{\lambda }_{j}x}\to f(x)e^{-i\lambda x}$ pointwise and every integrand is bounded in absolute value by the Lebesgue integrable function $|f|$. The Dominated convergence theorem therefore proves that $\hat{f}$ is continuous. In fact the Riemann-Lebesgue lemma also gives $\hat{f}(\lambda )\to 0$ as $|\lambda |\to \infty$.

Because $x^{-2}$ is integrable on $[1,\infty )$, the Dominated convergence theorem applied to the defining integral proves that $v$ is continuous on $\mathbb{R}$.

For $\lambda >0$, the change of variables formula $t=\lambda x$ givesand for $\lambda <0$ the analogous formula is obtained with $e^{it}$. On either open half-line the lower endpoint stays away from zero locally, so repeated differentiation under the integral sign proves smoothness. Thus

With $g=0$, pull the bounded measurable initial value back along the characteristic flow map:The flow is measurable and invertible, so $u$ is measurable and $\Vert u{\Vert }_{\infty }\le \Vert u_0{\Vert }_{\infty }$. Choose smooth $u_0^{(k)}$ converging to $u_0$ in $L_{\mathrm{loc}}^1$ with uniformly bounded essential suprema, and definePart a makes each $u^{(k)}$ a classical, hence weak, solution. On every compact subset of spacetime, the $C^1$ change-of-variables formula for the flow and its locally bounded Jacobian determinant give $u^{(k)}\to u$ in $L^1$. Passing to the limit in the weak identity by dominated convergence proves that $u$ is a bounded weak solution with initial datum $u_0$.