Suppose $f_n \geq 0$ for all $n \geq 1$, $f_n \to f$ a.e. on $[0, \infty)$ and there exists a constant $M>0$ such that $$ \sup\limits_{n} \int_{E} f_n(x)dx \leq M \mu(E)$$ for each measurable set $E \subset [0, \infty)$ with $\mu(E)>0$. Show that the set $A:=\{x \in [0,\infty):f(x)>M\}$ is of measure zero.

Here is my attempt:

We have that, for $A_m:=\{x \in [0,\infty):f(x) \geq M+ \frac{1}{m}\}$ for fixed $m$, by Chebychev's Inequality and Fatou's Lemma, \begin{align*} \left(M+\frac{1}{m}\right)\mu(A_m) &\leq \int_{A_m} f(x)dx \\ &\leq \lim\limits_{n \to \infty} \inf\limits_{k \geq n} \int_{A_m}f_k(x)dx \\ &\leq \lim\limits_{n \to \infty} \sup\limits_{k \geq n}\int_{A_m}f_k(x)dx \\ &\leq \sup\limits_{n} \int_{A_m} f_n(x)dx \leq M\mu(A_m)\end{align*}

Therefore, for all $m \in \mathbb{N}$, $(M+\frac{1}{m})\mu(A_m) \leq M \mu (A_m)$, and since $\mu$ is a nonnegative measure, this is only possible if $\mu(A_m)=0$ for all $m \in \mathbb{N}$.

Now notice $A=\{x:f(x)>M\}=\bigcup_{m=1}^{\infty}A_m$ since $$A_1 \subset A_2 \subset \ldots \subset A_m \subset A_{m+1} \subset \ldots$$ then by continuity of Lebesgue measure $$\mu(A)=\mu\left(\lim_{m\to \infty}A_m\right)=\lim_{m \to \infty} \mu (A_m)=0.$$

There are two obvious problems with my proof, and I can't seem to find a way to fix them:

(1) It is nowhere stated in the problem that $f_n$ are measurable.

(2) The $(A_m)$ described in my proof could also satisfy the above inequalities if $\mu(A_m)=\infty$. I am not sure what prevents them from being as such.

Any help would be appreciated! Thanks ahead of time.

  • 3
    $\begingroup$ (1) If they weren't measurable, $\int_E f_n(x)\,dx$ would not exist for $E$ with $\mu(E) > 0$. (2) Look at $A_m \cap [0,k]$ for $k \in \mathbb{N}\setminus\{0\}$. $\endgroup$ Jul 22, 2014 at 23:26
  • $\begingroup$ So it would suffice to assume $A_1$ is infinite in measure, say. Then set $B_k=A_1 \bigcap [0,k]$. But we would only have then that the supremum of the $\int_{B_k}f_n \leq$ Mk, right? That could still go to infinity. $\endgroup$
    – Darrin
    Jul 22, 2014 at 23:37
  • $\begingroup$ I think I have it. Suppose the latter were the case, i.e. suppose $\mu(A_m)=\infty$ for some $m \in \mathbb{N}$. Then for some $k \in \mathbb{N}$ we would have $k \geq \mu(A_m \cap [0,k])>0$. This means by the above reasoning that $(M+\frac{1}{m})\mu(A_m) \leq M \mu (A_m)$, $$(M+\frac{1}{m}) \leq M$$ which is impossible. Note also if $\mu(A)=\infty$ then because $f_n \to f$ we must have that $\mu(A_m)=\infty$ for some $m \in \mathbb{N}$ and the preceding counterargument still applies. $\endgroup$
    – Darrin
    Jul 23, 2014 at 5:16
  • $\begingroup$ I presume that you are using the Lebesgue measure? If so, just deal with bounded intervals in the domain and then you don't need to fiddle with sets of unbounded measure. $\endgroup$
    – copper.hat
    Jul 23, 2014 at 16:32

1 Answer 1


Let $A_n = f_n^{-1} (M,\infty)$. Let $B=A \cap [a,b]$ and $B_n = A_n \cap [a,b]$ where $0 \le a<b$.

Since $f_n(x) \to f(x)$ for ae. $x$, if $x \in B$, then $x \in B_n$ for all $n$ sufficiently large, that is $1_{B_n}(x) \to 1$. Hence $\mu(B \cap B_n) \to \mu B$.

We see that $f_n(x) \to f(x)$ for ae. $x \in B$, hence $\liminf_n f_n(x) 1_{B_n \cap B}(x) \ge f(x)1_B(x)$ for ae. $x$.

Fatou's lemma gives $\liminf_n \int f_n 1_{B_n \cap B} \ge \int \liminf f_n(x) 1_{B_n \cap B} \ge \int f 1_B = \int_B f$, and hence $\liminf_n \int f_n 1_{B_n \cap B} \ge \int_B f$.

Suppose $\mu B >0$. Then $\int_B f > M \mu B$, and hence $\liminf_n \int f_n 1_{B_n \cap B} > M \mu B$. Since $\mu(B \cap B_n) \to \mu B$, we see that we must have $\int f_n 1_{B_n \cap B} > M \mu(B \cap B_n)$ for some $n$.

It follows that $\mu B = 0$ and hence $\mu A = 0$.


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