= Solution
For the real-valued variables here, take a $\mathcal G$-measurable version of $X=\mathbb E[Y\mid\mathcal G]$. Since the variables are bounded, the <conditional expectation> defining identity extends to the bounded $\mathcal G$-measurable multiplier $X$, giving
$$
\mathbb E[XY]=\mathbb E\bigl[X\mathbb E[Y\mid\mathcal G]\bigr]=\mathbb EX^2.
$$
The <equality case for conditional second moments> now gives
$$
\mathbb E(Y-X)^2=\mathbb EY^2-2\mathbb E[XY]+\mathbb EX^2
=\mathbb EY^2-\mathbb EX^2=0.
$$
A nonnegative <random variable> with zero expectation vanishes with probability one. \b[Thus $X=Y$ as an <almost sure equality>.] The same proof works for square-integrable variables; boundedness is more than is needed.
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