Counter-example $L^p$ spaces

Question

Consider the following problem

Let $E$ a Lebesgue measurable set and let $f \in L^\infty(E)$. Is it true that is $f=0$ iff $\int_E fg = 0$ for all $g \in L^1(E)?$

I have already know that for the case $1 \leqslant p <\infty$ if $f \in L^p$ then $f=0$ iff $\int fg=0$ for all $g \in L^q$ for q conjugate of $p.$

The questions is, does it hold for $p=\infty?$ If yes, can I have some hint to prove? Or, can I have a counterexample?

Thank you everyone!

Answer 1

Define $A_n = E \cap [f > 0] \cap [-n,n]$ and $A = E\cap [f >0]$. Then, the indicator function $1_{A_n} \in L^{1}(E)$ for any $n \in \mathbb{N}$. The condition implies $$ \int_{E} f1_{A_n} = 0, \quad n \in \mathbb{N}$$ Note that $1_{A_n} \to 1_{A}$ monotonously, and thus by the monotone convergence theorem

$$\int_{E\cap[f>0]} f = \int_{E}\lim_{n\to\infty} f1_{A_n} = \lim_{n\to\infty} \int_{E} f 1_{A_n} = 0$$

Which shows $f \leq 0$ almost everywhere. Apply the same argument to $-f$ to obtain $f \geq 0$.

Answer 2

Assume the sigma algebra is $\{X,\emptyset\}$ with measure $\mu(X)=\infty$ and $\mu(\emptyset)=0$, then $L^{1}(X)=\{0\}$ but $f=\chi_{X}=1$ satisfies $\|f\|_{L^{\infty}}=1$.

However, the assertion is true with $\sigma$-finite measure space. Say, $X=\displaystyle\bigcup_{n}X_{n}$ with increasing $\{X_{n}\}$ and $\mu(X_{n})<\infty$, it is not hard to see that $\|f\|_{L^{\infty}(X)}=\lim_{n\rightarrow\infty}\|f\|_{L^{\infty}(X_{n})}$. For each $n$, setting $g_{n}=\chi_{\{f<0\}\cap X_{n}}$, then $\displaystyle\int_{X}fg_{n}d\mu=\displaystyle\int_{X_{n}}-f^{-}d\mu=0$. Similarly, we have $\displaystyle\int_{X_{n}}f^{+}d\mu=0$, so $f=0$ almost everywhere on $X_{n}$.