# Holder inequality with $q = \infty$

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 $$q=\infty.$$

• Do we define $P(F)$ to be the outer measure of $F$ with respect to measurable sets that contain $F$, i.e., $P(F) := \inf_{F' \in \mathscr{F}}{ P(F') }$? If so, then proving your statement (1) immediately implies the statement for all $F$. Commented Jan 1, 2020 at 19:06
• Also your notation $|1_F|^\infty$ is strange, and it seems like faulty reasoning. The $L^\infty$-norm of $|1_F|$ is $1$ which is unhelpful to proving your statement. Instead, you want to use the dominated convergence theorem. Commented Jan 1, 2020 at 19:10
• Thank you. I'll have a look at your answer in a second but what is wrong with this $$E[|X|1_F] = \left| \int_{\Omega}|X| 1_F dP \right| = \left(\int_{\Omega}|X| dP \right) ||1_F||^{\infty}$$?
– UBM
Commented Jan 1, 2020 at 19:17
• The problem is that it doesn't give you the estimate you want since $||1_F||^\infty = 1$ :). By the way, the notation $||1_F||_\infty$ is more common. Commented Jan 1, 2020 at 19:19
• oh, yes you are right, I was mistakenly assuming $P(F) = ||1_F||^{\infty}<\delta.$ I'll have a look at your answer then.
– UBM
Commented Jan 1, 2020 at 19:23

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.
• In case you're curious about my claim about there being such a subsequence $n_j$, let $n_j$ be such that $P(F_{n_j}) < 2^{-j}$. Commented Jan 1, 2020 at 19:16