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I'm trying to solve this question and I need help as to how to start. I realized that one direction is easy but the other, I have tried several times but it yields no results. The question is if $h \in L^+(X, M)$ and $ (X, M, \mu) $ is a measurable space. Let the function $\Lambda :M \rightarrow [0, \infty]$ defined by $ \Lambda (E) = \int _{E}h d \mu$ be a measure. Now, we need to proof that $ (X, M, \Lambda)$ is $\sigma$-finite if and only if $h< \infty$ $\mu$- almost everywhere.

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And which is the direction that you found easy, and which is the direction that you are having trouble with? –  Arturo Magidin Nov 16 '11 at 1:22
    
If $h \in {h} < \infty$ then $ (X, M, \mu ) $ is $\sigma $- finite. That direction is okay.. –  wright Nov 16 '11 at 1:27

1 Answer 1

In the comments you indicate that you ask:

If $(X,M,\Lambda)$ is $\sigma$-finite why is $h \lt \infty$ $\mu$-almost everywhere?

(Note the typo in your question: you wrote $\mu$ instead of $\Lambda$—according to your title $\mu$ is a finite measure already).

By hypothesis we can write $X = \bigcup_n E_n$ with $\Lambda(E_n) \lt \infty$ for all $n$.

Now $\Lambda(E_n) \lt \infty$ means that $$ \Lambda(E_n) = \int_{E_n} h\,d\mu = \int [E_n]\,h\,d\mu \lt \infty. $$ where I write $[E_n]$ for the characteristic function of $E_n$. This tells us that $[E_n]\,h$ is $\mu$-integrable. A non-negative $\mu$-integrable function is finite $\mu$-almost everywhere$\color{red}{(1)}$, so there is a $\mu$-null set $N_n$ such that $\{x\,:\,([E_n]h)(x) = \infty\} = \{x \in E_n\,:h(x) = \infty\} \subset N_n$.

Put $N = \bigcup_n N_n$ and note that $\mu(N) \leq \sum_n \mu(N_n) = 0$. Recall that $X = \bigcup_n E_n$, so $$\{x \in X\,:\,h(x) = \infty\} = \bigcup\nolimits_n \{x \in E_n\,:\,h(x) = \infty\} \subset \bigcup\nolimits_n N_n = N.$$ In other words, $\{x \in X\,:\,h(x)=\infty\}$ is a $\mu$-null set, so $h(x) \lt \infty$ almost everywhere.


$\color{red}{(1)}$A $\mu$-integrable function $f \geq 0$ is finite almost everywhere: Suppose not. Then the set $\{f = \infty\}$ has positive $\mu$-measure. Thus, $$\int f\,d\mu \geq \int_{\{f = \infty\}} f\,d\mu = \infty,$$ so $f$ is cannot be $\mu$-integrable, a contradiction.

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@ t.b, thanks for pointing it out to me... –  wright Nov 16 '11 at 8:15

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