Holder inequality with $q = \infty$

Question

Assume we have a probability space $(\Omega, \mathscr{F},P)$. As a part of a proof I found the following:

If $X \in L^1$, then for any $\epsilon >0$, there exists some $\delta > 0$ such that $$ P(F)<\delta \implies E[|X|1_F]< \epsilon.$$

I have two questions:

1) They don't say anything about the set $F.$ Should I interpret the question above as the following?

If $X \in L^1$, then for any $\epsilon >0$, there exists some $\delta > 0$ such that for all $F \in \mathscr{F}$ $$ P(F)<\delta \implies E[|X|1_F]< \epsilon.$$

2) Can I use Holder inequality with $q= \infty$ to prove 1)? Is correct the following?

Since $X \in L^1$ we have $\int_{\Omega}|X|dP = C$ for some $C \in \mathbb{R}^+.$ Also, by hypothesis, $P(F)= \int_{\Omega} 1_F dP < \delta.$ So choosing $\delta_{\epsilon} < \frac{\epsilon}{C}$ we have

\begin{align*} E[|X|1_F] &= \left| \int_{\Omega}|X| 1_F dP \right| \\ &\leq \int_{\Omega}|X| dP \int_{\Omega} |1_F|^{\infty} dP \tag*{(*)} \\ &= C \delta \\ &< \epsilon. \end{align*} Where in step (*) I've used Holder inequality with $p=1$ and $q=\infty.$

Answer 1

Usually this lemma is not proven by Holder inequality, but dominated convergence theorem. Suppose the claim (1) is false. Then for some $ \varepsilon > 0 $ there exists a sequence $ F_n $ with $ P(F_n) \rightarrow 0 $ but $ E[X|1_{F_n}] > \varepsilon $. But on the other hand, there is a subsequence $ f_{n_j} := X \cdot 1_{F_{n_j}} $ that is dominated by the integrable function $ |X| $ and converging to zero almost everywhere. The dominated convergence theorem states that $ E[X|1_{F_{n_j}}] \rightarrow 0 $, contradicting our earlier statement that these numbers are greater than $ \varepsilon $. There is probably a direct proof of this, but this was what came to mind, so I wrote it down :) Hope it helps.