7
$\begingroup$

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!

$\endgroup$

2 Answers 2

7
$\begingroup$

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 $$\mu(X)=\mu(\cup_{m=1}^{\infty} X_m)\leq \sum_{m=1}^{\infty}\mu(X_m)=0$$

$\endgroup$
3
  • $\begingroup$ Does this argument work even if $\mu(X) = \infty$? $\endgroup$
    – layman
    Aug 14, 2015 at 23:25
  • 2
    $\begingroup$ @user46944 $[0, \infty)$ is $\sigma$ finite... $\endgroup$
    – user251257
    Aug 15, 2015 at 0:03
  • 2
    $\begingroup$ Well just show $\mu(\{f>M\}\cap [-a,a] )= 0$ for each $a>0.$ $\endgroup$
    – zhw.
    Aug 15, 2015 at 0:04
3
$\begingroup$

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

$\endgroup$
2
  • $\begingroup$ Thanks, @zhw, but I am still a little confused: I tried using Fatou's lemma earlier, but I'm not sure how it helps in this situation. Why would this imply that $\mu\{x\in[0,\infty)|f(x)>M\}=0$? Am I missing something else? $\endgroup$ Aug 14, 2015 at 22:07
  • $\begingroup$ Yes, you're missing the second part of the problem. You now need to argue from $\int_E f\,d\mu \le M\mu (E)$ to the conclusion. $\endgroup$
    – zhw.
    Aug 14, 2015 at 23:49

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.