So I am to prove that for a measure space $(\Omega, \mathcal{A}, \mu)$ and a nonnegative, integrable function $f: \Omega \to \mathbb{\overline{R}}$, for all $\epsilon > 0$ there exists $A\in\mathcal{A}$ satisfying $\mu(A) < \infty$ such that

\begin{equation} \int_\Omega f \ d\mu \leqslant \int_A f \ d\mu + \epsilon. \end{equation}

My current approach:

Let $f_n: \Omega \to \mathbb{\overline{R}}$, $n \in \mathbb{N}$, be a sequence of nonnegative, measurable, simple functions such that $f_n \uparrow f$, and hence $\int_\Omega f_n \ d\mu \uparrow \int_\Omega f \ d\mu $. For any $k \in \mathbb{N}$, denote the $r_k$ values that $f_k$ takes on by $\alpha_m$ for $m \in \{1,\ldots, r_k \}$. Define the disjoint sets $A_m$ as

\begin{equation} A_m = f^{-1}(\alpha_m) \end{equation}

for each $m \in \{1,\ldots, r_k \}$. The integrability of $f$ insists that $\int_\Omega f \ d\mu < \infty$, hence also $\int_\Omega f_k \ d\mu < \infty$. Consequently, $f_k(A_m) = 0$ when $\mu(A_m) = \infty$. Let

\begin{equation} E_k = \bigcup_{m\in I} A_m, \quad I = \{m: \mu(A_m) < \infty\} \end{equation}

then clearly $\int_\Omega f_k \ d\mu = \int_{E_k} f_k \ d\mu $, and since $k$ was chosen arbitrarily,

\begin{equation} \int_{E_n} f_n \ d\mu \uparrow \int_\Omega f \ d\mu. \end{equation}

This implies that for all $\epsilon>0$, there exists $N \in \mathbb{N}$ such that $n>N$ implies

\begin{equation} \left|\int_{E_n} f_n \ d\mu -\int_\Omega f \ d\mu\right| = \int_\Omega f \ d\mu - \int_{E_n} f_n \ d\mu < \epsilon, \end{equation}


\begin{equation} \int_\Omega f \ d\mu \leqslant \int_{E_n} f_n \ d\mu + \epsilon, \quad \forall n \in \mathbb{N}, \end{equation}

and furthermore

\begin{equation} \int_\Omega f \ d\mu \leqslant \lim_{n\to\infty}\int_{E_n} f_n \ d\mu + \epsilon = \int_E f \ d\mu + \epsilon, \end{equation}

where $\lim_{n\to\infty} E_n = E$. Now it remains to show that $\mu(E) < \infty$. I know that it holds for $E_k$ for each $k \in \mathbb{N}$, since it is the finite union of (disjoint) sets of finite measure, but how do I prove that this also holds in the limit?

  • $\begingroup$ This will not be true for $E$, but this will be true for a subset (with finite measure) of $E$ that would be "large enough". This holds because the integral of a nonnegative integrable function is a finite measure (cf. my answer). $\endgroup$ – Julien Oct 15 '17 at 10:41

You can use that $E=\{x \in \Omega; f(x) \neq 0\}$ is $\mu-$finite, which allows you to build an increasing sequence of subsets $A_{n}$ (with $\mu(A_{n})<+\infty$) such that $A_{n} \rightarrow E$.

But then $\int_{A_{n}}f d\mu \rightarrow \int_{\Omega}f d\mu$ since the integral of $f$ (which is integrable and nonnegative) is a finite measure. But then you can find $N \in \mathbb{N}$ such that the required inequality holds for every $n \geq N$.


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