This is based on Durrett's 5.1.3
Prove Chebyshev's inequality. If $a > 0$ then $$\mathbb{P}(\lvert X \rvert \geq a | \mathcal{F}) \leq a^{-2}\mathbb{E}(X^2 | \mathcal{F})$$
First, I need to establish $X^2 \in L^1(\Omega, \Sigma, \mathbb{P})$, so the inequality is possible to have any meaning (otherwise functions are not defined). And i suppose $X \in L^1$, so the left side is defined.
But, following $L^1 \subseteq L^2$? I can't deduce anything about $X^2$.
Should I just suppose $X^2 \in L^1(\Omega, \Sigma, \mathbb{P})$? Or Durrett works in $L^2$?
We just solve problems from the book, so did during my fast-forward search I missed this assumption?
$\newcommand{\E}{\mathbb E}\newcommand{\P}{\mathbb P}$I think as long as one talks about the conditional expectation, then one assumes indirectly the existence of it. Otherwise the question would not make sense. Notice that we don't need the integrability of $X$ to define $\P(|X|\geq a \mid\mathcal F)$. We need the integrability of $\mathbf{1}_{|X|\geq a}$ which is trivially integrable.
Notice that: \begin{align} a\mathbf{1}_{|X|\geq a}\leq |X| \end{align} Squaring preserves the inequality since both sides are positive: \begin{align} a^2\mathbf{1}_{|X|\geq a}\leq |X|^2=X^2 \end{align} where $(\mathbf{1}_{|X|\geq a})^2=\mathbf{1}_{|X|\geq a}$ is used. By monotonicity and the linearity of the conditional expectation we have: \begin{align} a^2\E[\mathbf{1}_{|X|\geq a} \mid\mathcal F]\leq \E[X^2\mid \mathcal F] \end{align} Dividing both sides with $a^2$ yields: \begin{align} \P(|X|\geq a\mid\mathcal F)\leq a^{-2}\E[X^2\mid\mathcal F] \end{align}