A nonconstant centered integrable IID random walk is unbounded above

Let $M={\sup }_{n\ge 0}{S}_n$, a measurable $[0,+\mathrm{\infty }]$-valued random variable. For each $m\ge 1$, removing the first $m$ increments changes later partial sums only by the finite number $S_m$, and the removed partial sums form a finite set. Consequently $\{M=+\mathrm{\infty }\}=\{{\sup }_{k\ge 0}{\sum }_{i=m+1}^{m+k}{X}_i=+\mathrm{\infty }\}$ belongs to $\sigma (X_{m+1},X_{m+2},\dots )$. It is a tail event, so its probability is zero or one.

Suppose its probability were zero. Then $M$ is finite almost surely. Define $M^{\mathrm{\prime }}={\sup }_{k\ge 0}{\sum }_{i=2}^{k+1}{X}_i$, which is also finite almost surely and has the same law as $M$: for every real $b$, the event that the supremum is at most $b$ is the decreasing intersection of the corresponding finite-prefix events, whose probabilities agree by the IID hypothesis. The variable $M^{\mathrm{\prime }}$ is independent of $X_1$ by independence of their coordinate sigma-algebras. Decomposing the supremum according to whether the empty partial sum is used gives $M=\max (0,X_1+M^{\mathrm{\prime }})$ almost surely. On null exceptional sets replace $M$ and $M^{\mathrm{\prime }}$ by zero if needed, without changing their laws or these almost-sure assertions.

Put $A=\max (0,M^{\mathrm{\prime }}+X_1)$ and $B=M^{\mathrm{\prime }}$, so $A$ and $B$ are finite nonnegative variables with the same law. The map $u\mapsto \max (0,u)$ is one-Lipschitz and $B\ge 0$, hence $\mathrm{\mid }A-B\mathrm{\mid }\le \mathrm{\mid }{X}_1\mathrm{\mid }$. For each positive integer $K$, equality in law gives $\mathbb{E}[\min (A,K)-\min (B,K)]=0$. The difference converges almost surely to $A-B$ and has absolute value at most $\mathrm{\mid }A-B\mathrm{\mid }\le \mathrm{\mid }{X}_1\mathrm{\mid }$. Dominated convergence therefore yields $\mathbb{E}(A-B)=0$, without requiring either $A$ or $B$ to have finite expectation.

The identity $A-B=X_1+R$, where $R=(-B-X_1{)}^{+}$ and $u^{+}=\max (u,0)$, gives $0\le R\le \mathrm{\mid }{X}_1\mathrm{\mid }$. Centering and step 3.1 imply $\mathbb{E}R=0$, so $R=0$ almost surely. Thus $B+X_1=A$ almost surely and $B+X_1$ has the same law as $B$. Independence now gives ${\phi }_B(t){\phi }_{X_1}(t)={\phi }_B(t)$ for every real $t$.

Continuity and ${\phi }_B(0)=1$ supply $\delta >0$ such that ${\phi }_B(t)\mathrm{\ne }0$ for $\mathrm{\mid }t\mathrm{\mid }<\delta$. Step 4.1 then gives ${\phi }_{X_1}(t)=1$ on this interval. Choose a positive integer $q$ with $1\mathrm{/}q<\delta$. For every integer $k\ge q$, taking real parts gives $\mathbb{E}[1-\cos (X_1\mathrm{/}k)]=0$, so $\cos (X_1\mathrm{/}k)=1$ almost surely. Intersecting these countably many probability-one events, fix a resulting value $x$ of $X_1$. If $x\mathrm{\ne }0$, some $k\ge q$ has $0<\mathrm{\mid }x\mathrm{\mid }\mathrm{/}k<2\pi$, contradicting $\cos (x\mathrm{/}k)=1$. Therefore $X_1=0$ almost surely, contrary to the hypothesis.

The zero-probability alternative in step 1.1 is impossible, so $M=+\mathrm{\infty }$ almost surely. Each finite prefix has finite maximum, and hence removing it preserves an infinite supremum. Thus the limsup of the partial sums is also $+\mathrm{\infty }$ almost surely.