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The partial sums $s_n={\sum }_{k=1}^n{a}_k$ are nonnegative measurable functions and increase to an extended-valued function $s$. By the monotone convergence theorem,A nonnegative function with finite Lebesgue integral cannot equal $+\infty$ on a set of positive measure: otherwise its integral would exceed $M$ times that measure for every $M>0$. Thus the series converges to a finite value almost everywhere. Redefine it to be zero on the exceptional null set if a finite-valued representative is desired. The resulting function is Lebesgue integrable, with

Put $x=e^{-t}$. Thenwhere the second substitution is $u=(k+1)t$ and the final integral is the given gamma function value. The formula also holds for $k=0$, with value $1$.

For $0<x<1$, the exponential function expansion givesEvery term is nonnegative. By the monotone convergence theorem, integration can be interchanged with summation, and henceThis identity is one form of Sophomore's dream.

The monotone convergence theorem states that if measurable functions $0\le f_n\uparrow f$ pointwise on a measure space, then $\int f_n\uparrow \int f$, allowing the limit to be infinite.

Extend each restriction to $I_n$ by zero on $I\setminus I_n$. The functions $f{1}_{I_n}$ are nonnegative and increase pointwise to $f$ because the intervals are increasing and exhaust $I$. HenceFor nonnegative $f$, integrability means finiteness of this integral, soNo boundedness or closedness of the intervals is required.

For $\alpha >0$, integration by parts on $[0,R]$ givesThe exponential remainder tends to zero. Since the integrand is nonnegative, the monotone convergence theorem applied to truncations identifies the improper integral with the Lebesgue integral. ThereforeIts finite value proves integrability as well as evaluating it.

The monotone convergence theorem states that if $0\le h_n\uparrow h$ pointwise for measurable functions, then $h$ is measurable and $\int h={\lim }_n\int h_n$, allowing infinity.

Apply it to the partial sums $S_N={\sum }_{n=0}^N{f}_n$. They are nonnegative, measurable, and increase pointwise to $f$. Finite additivity of the integral and monotone convergence yieldThe assumed finite sum of integrals makes $f$ integrable. This proves the series-integration rule needed below directly from the stated theorem, without assuming an additional interchange theorem.

For $\alpha >0$, integration by parts on $[0,R]$ givesThe exponential remainder tends to zero. Since the integrand is nonnegative, the monotone convergence theorem applied to truncations identifies the improper integral with the Lebesgue integral. ThereforeIts finite value proves integrability as well as evaluating it.

The monotone convergence theorem states that if $0\le h_n\uparrow h$ pointwise for measurable functions, then $h$ is measurable and $\int h={\lim }_n\int h_n$, allowing infinity.

Apply it to the partial sums $S_N={\sum }_{n=0}^N{f}_n$. They are nonnegative, measurable, and increase pointwise to $f$. Finite additivity of the integral and monotone convergence yieldThe assumed finite sum of integrals makes $f$ integrable. This proves the series-integration rule needed below directly from the stated theorem, without assuming an additional interchange theorem.

First identify the recurrence type, rather than infer an occupation limit from a one-time probability. At time $k(a+b)$, return to zero requires exactly $kb$ upward steps. The binomial distribution giveswhere the last step is Stirling's approximation. At all other positive times the return probability is zero. Hence ${\sum }_{n\ge 0}{\mathbb{P}}_0(S_n=0)=\infty$. The recurrence criterion by return probabilities proves the recurrent property; by the irreducible Markov chain property established in (i), every state is recurrent. The equivalence of positive recurrence and existence of a stationary distribution for an irreducible Markov chain on a countable state space, together with (iii), makes this chain null recurrent.

For completeness, the zero occupation proportion follows directly by treating returns as a renewal process. Let $L_1,L_2,\dots$ be successive strictly positive return-cycle lengths to zero. The Strong Markov property makes them independent and identically distributed, the recurrent property makes them finite almost surely, and the null recurrent property gives $\mathbb{E}{L}_1=\infty$. For each positive integer $K$, the strong law of large numbers applied to $\min (L_j,K)$ givesTake a common event of probability one for these countably many $K$. By the monotone convergence theorem, the right side increases to infinity. Thus the number of completed return cycles by time $n$, divided by $n$, tends to zero. Counting the initial visit changes this ratio by at most $1/n$. Consequently,The walk visits zero infinitely often, but the asymptotic fraction of time spent there is zero.

The bounded-time optional sampling theorem says that for an integrable discrete-time martingale $(U_n)$ and a bounded stopping time $S$, $\mathbb{E}[U_S]=\mathbb{E}[U_0]$. It requires no assumption about an eventual unbounded stopping time.

Apply this theorem with $S=n\wedge \tau$, or simply take the expectation of the stopped martingale in part (ii). For $N>0$ we obtainSince $P_{n\wedge \tau }\ge 0$,The monotone convergence theorem now yields $\mathbb{E}[\tau ]\le 3abc/N<\infty$. In particular $\tau <\infty$ almost surely, which has been established without assuming that optional stopping is valid at $\tau$.

For each sample path with finite $\tau$, $P_{n\wedge \tau }$ eventually equals $P_{\tau }=0$. Also $0\le P_{n\wedge \tau }\le N^3$. The dominated convergence theorem gives $\mathbb{E}[P_{n\wedge \tau }]\to 0$. Taking the limit of a sequence in the earlier identity therefore provesFor example, initial holdings $(1,1,1)$ give expectation one, as the first transfer must eliminate its giver. This computes the first elimination time, rather than the later time when one player owns all tokens.

The monotone convergence theorem states that if measurable functions $f_n$ are nonnegative and $f_n\uparrow f$ pointwise, then $f$ is measurable andallowing either side to be infinite. The same conclusion holds under almost-everywhere monotone convergence after changing values on a null set. No assumption that the integrals stay bounded is required.

The dominated convergence theorem states: if measurable $f_n$ converge almost everywhere to $f$ and $|f_n|\le g$ almost everywhere for an integrable $g\ge 0$, then $f$ is integrable and $\int f_n\to \int f$. In fact $\int |f_n-f|\to 0$.

To derive it from the monotone convergence theorem, first prove Fatou's lemma. For nonnegative measurable $h_n$, the functions $k_N={\inf }_{n\ge N}{h}_n$ increase to $\mathrm{liminf}{h}_n$ and satisfy $\int k_N\le {\inf }_{n\ge N}\int h_n$. Monotone convergence yields $\int \mathrm{liminf}{h}_n\le \mathrm{liminf}\int h_n$.

Now $|f|\le g$ almost everywhere, so $f$ is integrable. Apply Fatou's lemma to the nonnegative functions $2g-|f_n-f|$, which converge almost everywhere to $2g$:Hence $\int |f_n-f|\to 0$, and $|\int f_n-\int f|\le \int |f_n-f|$ proves the assertion. Common exceptional null sets may be removed at the beginning since the sequence is countable.

The monotone convergence theorem states that if nonnegative measurable functions $h_n$ increase almost everywhere to $h$, then $\int h={\lim }_n\int h_n$, allowing infinity. The dominated convergence theorem states that if measurable $h_n\to h$ almost everywhere and $|h_n|\le g$ for an integrable $g$, then $h$ is integrable and $\int h_n\to \int h$.

The monotone convergence theorem states that for measurable functions $0\le f_1\le f_2\le \cdots$ converging pointwise to $f$, the limit is measurable andBoth sides may be infinite. Monotonicity and nonnegativity replace any need for an integrable dominating function.

The partial sums $S_n={\sum }_{k=1}^n{a}_k$ are nonnegative measurable functions and increase pointwise to the extended-valued sum $S={\sum }_{k=1}^{\infty }{a}_k$. The monotone convergence theorem and finite linearity giveA nonnegative measurable function with finite integral is finite almost everywhere; if it were infinite on a positive-measure set, its integral would be infinite. Define it arbitrarily on the null exceptional set if a real-valued representative is desired. Hence the series converges almost everywhere to an integrable function and

Substitute $x=e^{-t}$ to getwhere the second substitution $u=(k+1)t$ reduces the integral to the supplied gamma integral.

On $(0,1)$, $-x\log x\ge 0$. Expand $x^{-x}=e^{-x\log x}={\sum }_{j=0}^{\infty }(-x\log x{)}^j/j!$. The partial sums increase, so the monotone convergence theorem permits termwise integration. The integral of the $j$-th term is $1/(j+1{)}^{j+1}$, including $j=0$. Reindexing givesThis is finite, for instance because the summands with $k\ge 2$ are bounded by $2^{-k}$.

The Bessel process is stopped in the compact interval $[\alpha ,\beta ]$, where its drift and diffusion coefficients are bounded. In particular its stopped stochastic integral has expectation zero on every finite time interval.

For $k>1$, taking expectations in the stopped equation givesSince $Y_t\le \beta$, this bounds $\mathbb{E}(t\wedge \tau )$ uniformly in $t$. For $k<1$, the same argument with $1-\mathbb{E}{Y}_t$ gives $\mathbb{E}(t\wedge \tau )\le 2\beta (1-\alpha )/(1-k)$.

If $k=1$, the stopped process is a stopped Brownian motion. Itô's formula gives $\mathbb{E}(Y_t-1{)}^2=\mathbb{E}(t\wedge \tau )$, which is at most $\max \{(1-\alpha {)}^2,(\beta -1{)}^2\}$. Thus in every case the monotone convergence theorem gives $\mathbb{E}\tau <\infty$, so $\tau <\infty$ almost surely. By path continuity its value at exit is one of the two endpoints. They are disjoint possibilities, hence

Let $K={\sup }_{x}f(x)<\infty$ and let $C$ be a Lipschitz constant for $f$. Sinceboth coefficients $f$ and $-f^2/2$ are bounded and globally Lipschitz. The global existence theorem for stochastic differential equations with Lipschitz coefficients gives a unique global strong solution of a stochastic differential equation.

The same Itô formula cancellation as before givesFor $S=T_{-a}\wedge T_b$, the stopped exponential is bounded between $e^{-a}$ and $e^b$. Continuity and strict positivity of $f$ giveThe Itô isometry therefore yieldsThe monotone convergence theorem proves $\mathbb{E}S<\infty$. The bounded optional stopping argument is now identical to part (b)(iii), and givesPassing from level hitting to strict exceedances as in part (b)(iv),No global positive lower bound on $f$ was needed: compact-interval positivity suffices for each exit argument. The law is independent of the particular $f$ because the scale function is always $s(x)=e^x$: its infinitesimal generator satisfiesThis is the exponential scale function for a drift tied to diffusivity. Changing $f$ changes how quickly the path is traversed, but leaves these boundary probabilities and the maximum law unchanged.

For a submartingale with ${\sup }_n\mathbb{E}|X_n|=C<\infty$, its positive part is a submartingale by the conditional Jensen inequality. Thus the displayed conditional expectations increase in $p$. Their limits are finite almost surely and have expectations at most $C$, by the monotone convergence theorem. Conditional monotone convergence and the tower property of conditional expectation give $\mathbb{E}[M_{n+1}\mid {\mathcal{F}}_n]=M_n$. Moreover $M_n\ge X_n^{+}$. Consequently $Y_n=M_n-X_n$ is a nonnegative supermartingale, ${\sup }_n\mathbb{E}{Y}_n\le 2C$, and $X=M-Y$. The resulting martingale need not have uniform integrability.

Suppose bounded $V\ge 0$ is bounded below by a positive constant on some nonempty open interval, and $f\ge C>0$. The Feynman-Kac formula with a bounded potential gives $u(t,x)\ge C{\mathbb{E}}_x{e}^{{\int }_0^{t}V(B_s)ds}$. Infinite occupation time of one-dimensional Brownian motion makes that integral tend to infinity almost surely. The monotone convergence theorem then gives $u(t,x)\to \infty$ for every fixed $x$. Thus no globally bounded classical solution exists under these hypotheses, although boundedness on every finite time slab is compatible with the conclusion.

Three players with positive integer holdings exchange a token by choosing an unordered active pair uniformly, then choosing the transfer direction fairly. Let $N$ be the conserved total, and $\tau$ the stopping time of the first elimination. Given holdings $(x,y,z)$ before $\tau$, the conditional expectation of the change of their product is $-(x+y+z)/3=-N/3$: transfers in each pair decrease the mean product by the third holding. Thus the displayed stopped process is a martingale. Its nonnegative product term gives $\mathbb{E}[n\wedge \tau ]\le 3abc/N$, so the monotone convergence theorem first proves integrability of $\tau$. The bounded product then vanishes by the dominated convergence theorem, yielding $\mathbb{E}[\tau ]=3abc/N$. Continuing the deterministic time term beyond elimination instead produces a submartingale.