Proof of Almost Sure Equality Preserves Square-Integrability and the Mean-Square Norm

Let $A'\in\mathcal{F}$ be an event with $P(A')=0$ such that $X(\omega)=Y(\omega)$ for every $\omega\in\Omega\setminus A'$, as furnished by the hypothesis, and write $\mathbf{1}_{A'}$ for the function on $\Omega$ equal to $1$ on $A'$ and $0$ off $A'$.

Step 1 (the exceptional part has expectation zero). The map $Y^{2}\mathbf{1}_{A'}$ is a nonnegative random variable vanishing at every point outside $A'$. Every nonnegative simple function bounded above by it therefore vanishes off $A'$, so each of its values other than $0$ is taken on a subset of $A'$, an event of probability $0$, and its integral is $0$. By the definition of the integral of a nonnegative random variable as the least upper bound of those integrals, $\mathbb{E}[Y^{2}\mathbf{1}_{A'}]=0$.

Step 2 (domination). Pointwise $Y^{2}\le X^{2}+Y^{2}\mathbf{1}_{A'}$: at a point of $\Omega\setminus A'$ one has $Y=X$, so $Y^{2}=X^{2}$ and the second summand is $0$; at a point of $A'$ the second summand alone equals $Y^{2}$, and the first is nonnegative.

Step 3 (conclusion). By the additivity and monotonicity of the integral of nonnegative random variables in the linearity and monotonicity theorem, Steps 1 and 2 give

so $Y$ is square-integrable. The hypothesis is symmetric in $X$ and $Y$, and $Y$ has just been shown square-integrable, so the same argument with the roles exchanged gives $\mathbb{E}[X^{2}]\le\mathbb{E}[Y^{2}]$. Hence $\mathbb{E}[X^{2}]=\mathbb{E}[Y^{2}]$, and since the mean-square norm of a square-integrable random variable is by definition determined by the expectation of its square, $X$ and $Y$ have the same mean-square norm. $\square$