# If$\mu$ is $\sigma$- finite, $\epsilon>0$, there exists $A\in \mathcal{A}$ such that $\mu(A)<\infty$ and $\epsilon+\int_A f>\int f$

Problem

Let $X\mathcal{A},\mu$ be a $\sigma$-finite measure space. Suppose $f$ is non-negative and integrable. Prove that if $\epsilon>0$, there exists $A\in \mathcal{A}$ such that $\mu(A)<\infty$ and $$\epsilon+\int_A f \ d\mu>\int f \ d\mu.$$

Attempt

First, I will prove two propositions.

Proposition 1: If $f$ is non-negative measurable function, then $$\lim_{n\to \infty} \int \min(f,n) \rightarrow \int f$$

Proof: Note that $\min(f,n)$ increases to $f$ so the result follows from the monotone convergence theorem.

Proposition 2: Suppose $\mu$ is $\sigma$-finite and $f$ is integrable. Then given $\epsilon>$, there exists $\delta$ such that $$\int_A |f(x)| \ d\mu < \epsilon,$$ whenever $\mu(A)<\delta$.

Proof: Given $\epsilon>0$, from proposition $1$ we may find $n_0$ such that $\int_A |f(x)| \ d\mu - \int_A \min(|f(x)|,n_0) \ d\mu<\frac{\epsilon}{2}.$ Take $\delta<\frac{\epsilon}{2n_0}$. then: \begin{align*} \int_A |f(x)|&=\int_A |f(x)| -\min(|f(x)|,n_0)\ d\mu+\int_A \min(|f(x)|,n_0)\ d\mu \\ &< \frac{\epsilon}{2}+n\mu(A) \\ &<\epsilon, \end{align*} as soon as $\mu(A)<\delta$.

Finally we prove the result. $\epsilon>0$, use proposition $2$ to insure that $\int_B f\ d\mu <\epsilon$ whenever $\mu(B)<\delta$. Let $B^c=A$. Then \begin{align*} \int f \ f\mu&=\int_A f\ d\mu+\int_B f\ d\mu \\ &<\int_A f \ f\mu+\epsilon, \end{align*} as desired.

Concerns

For proposition $2$ as well as for the problem, I did not use $\sigma$-finiteness. I cannot see anything wrong with proposition $2$. However, in my solution to the problem, I have not guaranteed that such sets (i.e $B$ such that $\mu(B)<\delta$) actually exist. Please let me know what I did wrong here.

• This is true without assuming sigma-finiteness. A bigger problem with your solution than the one you point out is that you've given no reason to think that $\mu(A)<\infty$, and in fact if $\mu$ is infinite your construction guarantees $\mu(A)=\infty$. You might start over, approximating $f$ in $L^1$ by a simple function... Commented Jun 24, 2015 at 2:02
• @DavidC.Ullrich Thank-you for your response. I started the problem over and put my work below. If you have time, could you let me know how I did? Commented Jun 24, 2015 at 5:06

Since $\mu$ is $\sigma$-finite, we may find a countable sequence $(E_k)$ with $\mu(E_k)<\infty$ for all $k$ so that $X=\bigcup E_k$. Let $A_n=\bigcup_{k=1}^{n} E_k$. Then $\bigcup A_n$, $\mu(A_n)<\infty$ and $A_n \subset A_{n+1}$. Since $f$ is integrable, we may find an increasing sequence $(g_n)$ of simple functions so that $\int g_n \rightarrow \int f$ and $g_n<f$ for all $n$. This also implies that $\int g_n \chi_{A_n} \rightarrow \int f$ and $g_n\chi_{A_n}<f\chi_{A_n}$. Thus, given $\epsilon>0$, we have for some $n_0$, $$\int f<\epsilon + \int g_{n_0} \chi_{A_{n_0}}<\epsilon+\int_{A_{n_0}} f.$$ Since $\mu(A_{n_0})<\infty,$ we have the result.
• Fine. But sigma-finite really has nothing to do with it! You don't need those $A_n$. Pick a simple function $g$ with $0\le g\le f$ so $\int g>-\epsilon+\int f$. Now just because $g$ is simple (and has finite integral) it follows that there is a set $A$ with $\mu(A)<\infty$ such that $g$ vanishes off $A$... Commented Jun 24, 2015 at 5:15
• Ahh I got this problem from a textbook so I was hung up on using $\sigma$- finiteness...Thank-you again for your response! Commented Jun 24, 2015 at 5:17
• although I do not see immediately what the set $A$ will be...will it be the union of all the sets on which the simple function takes on the value $1$? Commented Jun 24, 2015 at 5:19
• I suspect that what you meant there is correct. Say $g=\sum c_j\chi_{E_j}$. Then $A=\bigcup E_j$ works. Commented Jun 24, 2015 at 5:22