TheoremBase

Proof

Write 1D\mathbf{1}_D for the function equal to 11 on DD and 00 elsewhere, and N=Ω∖Ω∗N=\Omega\setminus\Omega_*. Since PP is a probability measure and P(Ω)=P(Ω∗)+P(N)P(\Omega)=P(\Omega_*)+P(N) by additivity, P(N)=0P(N)=0.

A null-integrand fact. If f:Ω→Rf:\Omega\to\mathbb{R} is a nonnegative random variable vanishing at every point of Ω∗\Omega_*, then E[f]=0\mathbb{E}[f]=0. Indeed, every simple function ss with 0≤s≤f0\le s\le f vanishes on Ω∗\Omega_*, so each set on which ss takes a nonzero value is contained in NN and therefore has probability 00 by monotonicity of the measure; hence the integral of ss is 00. Taking the supremum over such ss, which is the definition of the integral of a nonnegative measurable function, gives E[f]=0\mathbb{E}[f]=0.

Claim 1. The difference X−YX-Y is a random variable, being the composition of the sequentially continuous map (u,v)↦u−v(u,v)\mapsto u-v on R2\mathbb{R}^2 with the map having components XX and YY, by measurability of sequentially continuous functions of measurable Euclidean maps. Put Z=max⁡(X−Y,0)Z=\max(X-Y,0); this is a random variable, since for a real number cc the set {Z>c}\{Z>c\} equals Ω\Omega when c<0c<0 and equals {X−Y>c}\{X-Y>c\} when c≥0c\ge0.

At every point of Ω\Omega we have 0≤Z≤2K0\le Z\le2K, and at every point of Ω∗\Omega_* we have Z=0Z=0 by hypothesis. By the null-integrand fact, E[Z]=0\mathbb{E}[Z]=0. Since X−Y≤ZX-Y\le Z at every point of Ω\Omega, linearity and monotonicity of the integral give

E[X]−E[Y]=E[X−Y]≤E[Z]=0.\mathbb{E}[X]-\mathbb{E}[Y]=\mathbb{E}[X-Y]\le\mathbb{E}[Z]=0 .

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

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

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

First, E[U 1N]=0\mathbb{E}\big[U\,\mathbf{1}_{N}\big]=0: the two nonnegative random variables max⁡(U,0)1N\max(U,0)\mathbf{1}_N and max⁡(−U,0)1N\max(-U,0)\mathbf{1}_N vanish on Ω∗\Omega_*, so both have expectation 00 by the null-integrand fact, and their difference is U1NU\mathbf{1}_N. Hence E[U 1Ω∗]=E[U]\mathbb{E}[U\,\mathbf{1}_{\Omega_*}]=\mathbb{E}[U] by linearity.

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

E[W+]=E[W+1Ω∗]≤E[U 1Ω∗]=E[U],\mathbb{E}[W^+]=\mathbb{E}\big[W^+\mathbf{1}_{\Omega_*}\big]\le\mathbb{E}\big[U\,\mathbf{1}_{\Omega_*}\big]=\mathbb{E}[U],

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

Hence WW is integrable, with E[W]=E[W+]−E[W−]\mathbb{E}[W]=\mathbb{E}[W^+]-\mathbb{E}[W^-]. Both E[W+]\mathbb{E}[W^+] and E[W−]\mathbb{E}[W^-] lie in the interval from 00 to E[U]\mathbb{E}[U], so their difference has absolute value at most E[U]\mathbb{E}[U], that is ∣E[W]∣≤E[U]|\mathbb{E}[W]|\le\mathbb{E}[U]. ■\blacksquare

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