Proof of Almost Sure Inequalities Between Bounded Random Variables Pass to Expectations

Write $\mathbf{1}_D$ for the function equal to $1$ on $D$ and $0$ elsewhere, and $N=\Omega\setminus\Omega_*$. Since $P$ is a probability measure and $P(\Omega)=P(\Omega_*)+P(N)$ by additivity, $P(N)=0$.

A null-integrand fact. If $f:\Omega\to\mathbb{R}$ is a nonnegative random variable vanishing at every point of $\Omega_*$, then $\mathbb{E}[f]=0$. Indeed, every simple function $s$ with $0\le s\le f$ vanishes on $\Omega_*$, so each set on which $s$ takes a nonzero value is contained in $N$ and therefore has probability $0$ by monotonicity of the measure; hence the integral of $s$ is $0$. Taking the supremum over such $s$, which is the definition of the integral of a nonnegative measurable function, gives $\mathbb{E}[f]=0$.

Claim 1. The difference $X-Y$ is a random variable, being the composition of the sequentially continuous map $(u,v)\mapsto u-v$ on $\mathbb{R}^2$ with the map having components $X$ and $Y$, by measurability of sequentially continuous functions of measurable Euclidean maps. Put $Z=\max(X-Y,0)$; this is a random variable, since for a real number $c$ the set $\{Z>c\}$ equals $\Omega$ when $c<0$ and equals $\{X-Y>c\}$ when $c\ge0$.

At every point of $\Omega$ we have $0\le Z\le2K$, and at every point of $\Omega_*$ we have $Z=0$ by hypothesis. By the null-integrand fact, $\mathbb{E}[Z]=0$. Since $X-Y\le Z$ at every point of $\Omega$, linearity and monotonicity of the integral give

Claim 2. If $X=Y$ on $\Omega_*$, then both $X\le Y$ and $Y\le X$ hold on $\Omega_*$, so Claim 1 applied twice gives $\mathbb{E}[X]\le\mathbb{E}[Y]$ and $\mathbb{E}[Y]\le\mathbb{E}[X]$.

Claim 3. The constant functions with values $c$ and $-c$ are random variables bounded in absolute value by $\max(K,c)$, with expectations $c$ and $-c$. On $\Omega_*$ we have $X\le c$ and $-c\le X$, so Claim 1, applied with the bound $\max(K,c)$ in place of $K$, gives $\mathbb{E}[X]\le c$ and $-c\le\mathbb{E}[X]$, that is $|\mathbb{E}[X]|\le c$.

Claim 4. Put $W^+=\max(W,0)$ and $W^-=\max(-W,0)$, both random variables by the argument used for $Z$ in Claim 1, both nonnegative, with $W=W^+-W^-$ and $W^-\le K'$ at every point of $\Omega$.

First, $\mathbb{E}\big[U\,\mathbf{1}_{N}\big]=0$: the two nonnegative random variables $\max(U,0)\mathbf{1}_N$ and $\max(-U,0)\mathbf{1}_N$ vanish on $\Omega_*$, so both have expectation $0$ by the null-integrand fact, and their difference is $U\mathbf{1}_N$. Hence $\mathbb{E}[U\,\mathbf{1}_{\Omega_*}]=\mathbb{E}[U]$ by linearity.

Next, $W^+\mathbf{1}_N$ is nonnegative and vanishes on $\Omega_*$, so $\mathbb{E}[W^+\mathbf{1}_N]=0$ and therefore $\mathbb{E}[W^+]=\mathbb{E}[W^+\mathbf{1}_{\Omega_*}]$, both sides possibly infinite a priori. At every point of $\Omega$ we have $W^+\mathbf{1}_{\Omega_*}\le U\,\mathbf{1}_{\Omega_*}$: on $\Omega_*$ this is the hypothesis $|W|\le U$, and off $\Omega_*$ both sides vanish. Monotonicity of the integral therefore gives

which is finite. The same argument applied to $W^-$, which also satisfies $W^-\le U$ on $\Omega_*$, gives $\mathbb{E}[W^-]\le\mathbb{E}[U]$, and $\mathbb{E}[W^-]$ is finite in any case because $0\le W^-\le K'$.

Hence $W$ is integrable, with $\mathbb{E}[W]=\mathbb{E}[W^+]-\mathbb{E}[W^-]$. Both $\mathbb{E}[W^+]$ and $\mathbb{E}[W^-]$ lie in the interval from $0$ to $\mathbb{E}[U]$, so their difference has absolute value at most $\mathbb{E}[U]$, that is $|\mathbb{E}[W]|\le\mathbb{E}[U]$. $\blacksquare$