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The Strong Markov property at the first visit to $0$ gives the singleton equilibrium potential from the Green functionLet $U_n$ be an increasing exhaustion of $G$ by finite sets containing $0$, and let $g_{U_n}(x,0)$ be the Green function of a transient weighted graph stopped on leaving $U_n$. Each $g_{U_n}(\cdot ,0)$ has finite support. The Green identity and the Markov property give, for $m\ge n$,By the monotone convergence theorem, the right side tends to zero as $m,n\to \infty$, while $g_{U_n}(x,0)\uparrow g(x,0)$. Thus $g(\cdot ,0)$ is an $H_0$ limit of finitely supported functions. Dividing by the positive number $g(0,0)$ proves $h\in H_0$.

The inverse change of variables is $x=v/(1+u)$, $y=uv/(1+u)$. Thus the triangular region becomesThe upper limit depends on $u$: the new region is not a rectangle. The absolute inverse Jacobian determinant is $x^2/(x+y)$, so the change of variables formula cancels the rational factor in the integrand:Although the original integrand is undefined at the vertex $x=y=0$, its nonnegative improper integral exists. One can apply the change of variables formula on $x\ge \epsilon$ and let $\epsilon \downarrow 0$ by the monotone convergence theorem; the transformed integrand has the finite integral just computed.

At a fixed radius, Parseval's identity applied to the preceding Fourier series givesIn polar coordinates, $dA=rdrd\theta$. Integrating over radius and interchanging the nonnegative sum and integral givesThe interchange follows from the monotone convergence theorem. This is the special case of the area norm of the Poisson extension in the unit disk for the odd step boundary data.

For $x>0$ the geometric ratio $e^{-2x}$ has modulus less than one, andEvery summand is nonnegative. The monotone convergence theorem therefore permits integration of the sum:The integral is finite because its integrand is bounded above by $e^{-x}$, whose integral is one. The displayed paired sums are the even partial sums of the alternating harmonic series; its odd partial sums differ by a term tending to zero. Thus the required equality holds with the usual series ordering. One can additionally evaluate it by $t=e^{-x}$:Grouping into nonnegative pairs supplies the justified interchange; absolute convergence of the ungrouped integrated alternating series is not needed and would be false.

A real measurable function is Lebesgue integrable precisely when the integral of its absolute value is finite. For every integer $k\ge 1$, on $I_k=[k+1/6,k+5/6]$ one has $|\sin \pi x|\ge 1/2$ and $x\le k+1$. ConsequentlyThe intervals have disjoint interiors. Countable additivity of the nonnegative Lebesgue integral, or the monotone convergence theorem applied to their finite unions, givesThe harmonic series diverges, so the function is not Lebesgue integrable. Cancellation between adjacent oscillations cannot give absolute integrability.

The monotone convergence theorem states that if measurable nonnegative functions satisfy $f_n\uparrow f$ almost everywhere on a measure space, thenallowing either side to be $+\infty$.

The monotone convergence theorem says that if $0\le f_j\uparrow f$ almost everywhere on a measure space, then $\int f_j\uparrow \int f$, allowing infinity. The dominated convergence theorem says that if $g_j\to g$ almost everywhere and $|g_j|\le H$ for an integrable $H$, then $g$ is integrable and $\int g_j\to \int g$.

For $a\ne 0$, the truncated functions $f_a{1}_{[-j,j]}$ increase to $f_a$. Each has the elementary integral $2\arctan (|a|j)/|a|$. Thus monotone convergence givesFor fixed $x$, eventually $|x|\le n\pi /2$ and $n\sin (x/n)\to x$, so $g_n(x)\to 1/(1+x^2)$. Inside the support the supplied inequality yields $n^2{\sin }^2(x/n)\ge 4x^2/{\pi }^2$. Outside the support $g_n=0$. Consequently the global bound isand this majorant is integrable by the already proved formula with $a=2/\pi$. Dominated convergence now gives $\boxed{\int g_n\to \int (1+x^2{)}^{-1}dx=\pi }$. The majorant is independent of $n$, including on the moving support.

Put $G_N={\sum }_{n=1}^N|f_n|$. These are nonnegative measurable functions increasing to $G={\sum }_{n=1}^{\infty }|f_n|$, possibly initially with value infinity. Monotone convergence theorem givesAn infinite value on a set of positive measure would force an infinite integral. Thus $G$ is finite almost everywhere and, after setting it to zero on its null exceptional set if desired, is integrable.

Where $G$ is finite, the real series ${\sum }_n{f}_n$ is absolutely convergent. Define its sum to be $h$ there and zero on the exceptional null set. Measurability follows as an almost-everywhere limit of partial sums, and $|h|\le G$ almost everywhere gives integrability. Every partial sum is dominated by $G$, so dominated convergence theorem yieldsIn fact the convergence is also in the integrable-function norm:

The monotone convergence theorem says that if $0\le f_n\uparrow f$ almost everywhere and the functions are measurable on a measure space, thenwith infinite integrals permitted. The limit is measurable after the usual modification on any exceptional null set.

The dominated convergence theorem says that if measurable functions $f_n$ converge almost everywhere to $f$ and satisfy $|f_n|\le g$ almost everywhere for an integrable nonnegative function $g$, then $f$ is integrable andThe theorem applies equally to real or complex functions. In particular domination must be by one fixed Lebesgue integrable function, rather than by a different bound for each $n$.

The monotone convergence theorem states that if $0\le h_n\uparrow h$ pointwise almost everywhere and the functions are measurable, then $\int h_n\uparrow \int h$, allowing the value infinity.

Interpret the displayed densities as zero outside $(0,1)$, so no undefined logarithm is evaluated there. Each $f_k$ is nonnegative and measurable. With $x=e^{-t}$,where the last equality is integration by parts. This is finite, proving Lebesgue integrability.

The nonnegative partial sums satisfy, for $0<x<1$,They vanish outside the interval as well. Monotone convergence therefore givesThe series is finite by comparison with the convergent series ${\sum }_{j\ge 1}{j}^{-2}$, whose convergence follows from the integral test. Since $g\ge 0$, this finite integral proves absolute Lebesgue integrability as required.

The monotone convergence theorem says that if measurable functions satisfy $0\le f_1\le f_2\le \cdots$ and converge pointwise to $f$, then $f$ is measurable andallowing the common value to be $+\infty$.

The partial sums $S_N={\sum }_{k=0}^N{a}_k$ are measurable, nonnegative and increasing. By the monotone convergence theorem,A nonnegative measurable function with finite integral is finite almost everywhere: if it were infinite on a positive-measure set, every constant multiple of that set's indicator would be a lower bound, forcing an infinite integral. Thus the series converges almost everywhere.

The monotone convergence theorem says that if measurable functions satisfy $0\le f_1\le f_2\le \cdots$ and converge pointwise to $f$, then $f$ is measurable andallowing the common value to be $+\infty$.

The partial sums $S_N={\sum }_{k=0}^N{a}_k$ are measurable, nonnegative and increasing. By the monotone convergence theorem,A nonnegative measurable function with finite integral is finite almost everywhere: if it were infinite on a positive-measure set, every constant multiple of that set's indicator would be a lower bound, forcing an infinite integral. Thus the series converges almost everywhere.

The function is nonnegative and measurable, so Lebesgue integrability is equivalent to finiteness of its integral. For $\alpha \ne -1$,Letting $\epsilon \downarrow 0$ gives a finite limit exactly when $\alpha >-1$; at $\alpha =-1$ the integral is $-\log \epsilon$ and diverges. By the monotone convergence theorem for truncations,

For $\alpha \ne -1$,The limit as $R\to \infty$ is finite exactly when $\alpha <-1$; the case $\alpha =-1$ diverges logarithmically. Positivity and the monotone convergence theorem identify these limits with the Lebesgue integral. Hence

Apply the monotone convergence theorem to $s_N={\sum }_{k=1}^N|a_k|$. Its limit $s$ satisfiesThus $s$ is finite almost everywhere, which proves absolute convergence of ${\sum }_k{a}_k$ almost everywhere. Define its sum $a$ there, and set $a=0$ on the exceptional null set. Then $|a|\le s$, so $a$ is integrable. The partial sums of ${\sum }_k{a}_k$ are also dominated by $s$; dominated convergence givesThe series of integrals converges absolutely, since $|\int a_k|\le \int |a_k|$.

For $0<x<1$, expand $(1-x{)}^{-2}={\sum }_{k\ge 1}k{x}^{k-1}$. Multiplying by $(\log x{)}^2$ gives nonnegative terms, so the monotone convergence theorem permits termwise integration. Put $x=e^{-t}$ to obtainwhere scaling followed by two integrations by parts evaluates the last integral. HenceThe finite value also establishes integrability at both endpoints.

The monotone convergence theorem states that for nonnegative measurable functions $f_n\uparrow f$ almost everywhere, $\int f={\lim }_n\int f_n$, allowing the value $+\infty$.

The dominated convergence theorem states that if measurable functions $f_n\to f$ almost everywhere and $|f_n|\le g$ almost everywhere for some integrable $g$, then $f$ is integrable and $\int f_n\to \int f$. In fact $\int |f_n-f|\to 0$ as well. Both statements apply on any measure space.

The monotone convergence theorem states that if $f_n$ are nonnegative measurable functions on a measure space and $f_n\uparrow f$ almost everywhere, thenEither side is allowed to be $+\infty$; no integrable dominating function is required.