# Showing that a set is not infinite in measure

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.

• (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 $k \in \mathbb{N}\setminus\{0\}$. Jul 22, 2014 at 23:26
• 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. Jul 22, 2014 at 23:37
• 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. Jul 23, 2014 at 5:16
• 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. Jul 23, 2014 at 16:32

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 $\mu A = 0$.