If $\sup_n\int_E f_n(x)\ \mathsf dx\leq M\mu(E)$ then the measure of $\{x\in [0,\infty)\mid f(x)>M\}=0$.

Question

This question came up when I was studying for an analysis qualifying exam:

Suppose $f_n\geq 0$ for all $n\geq 1$, $f_n\rightarrow f$ a.e. on $[0,\infty)$ and there exists $M>0$ such that $$\sup_n\int_E f_n(x)\ \mathsf dx\leq M\mu(E)$$ for every measurable set $E\subset [0,\infty)$ with $\mu(E)>0$. Then $\mu\{x\in [0,\infty)\mid f(x)>M\}=0$. ($\mu$ denotes Lebesgue measure on $\mathbb{R}$.)

I have been trying to do something with Chebyshev's inequality, but I'm not sure I am on the right track. I would appreciate any pointers. Thanks in advance!

Answer 1

Let $X=\{x\in [0,\infty): f(x)>M\}$. Suppose $\mu(X)>0$. As @zhw. notes, Fatou's Lemma implies

$$\int_X f\leq \liminf_{n\to\infty}\int_X f_n\leq \limsup_{n\to\infty} \int_X f_n\leq M\mu(X)$$However, by definition, $\int_X f>M\int_X=M\mu(X)$. Combining these gives $M\mu(X)<M\mu(X)$, or $M<M$. Contradiction. So $\mu(X)=0$.

Edit: This is for the case $\mu(X)<\infty$.We need to treat the argument $\mu(X)=\infty$ separately.

If $f\in L^1$, then we don't have to worry about this, since $\int_X f>M\mu(X)$ implies $\int f=\infty$ which contradicts $f\in L^1$. Will work on a proof or counterexample for $f\not\in L^1$.

Edit 2: As mentioned in the comments, you can write $X=\cup_{m=1}^{\infty} X_m$ with $\mu(X_m)<\infty$. Just ignore those $X_m$ with measure $0$ and you can assume $0<\mu(X_m)<\infty$. Then the argument above still holds (since $X_m\subset X$, $f>M$ on $X_m$) implies that $\mu(X_m)=0$ for all $m$, so $$\mu(X)=\mu(\cup_{m=1}^{\infty} X_m)\leq \sum_{m=1}^{\infty}\mu(X_m)=0$$

Answer 2

Hint: Use Fatou's Lemma to see $\int_E f\,d\mu \le M\mu (E).$