0
$\begingroup$

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}

hence,

\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?

$\endgroup$
  • $\begingroup$ 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
1
$\begingroup$

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 $n \geq N$.

$\endgroup$

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.