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All sets described are finite subsets of $\mathbb{R}$.

Given a measurable and integrable non-negative function $f$ over some measurable domain $A$, I'm trying to show $f$ is integrable iff the collection $\lbrace \mu(A_k)\rbrace$ is summable where $A_k=\lbrace x\in E\,|n\le f(x) \rbrace.$

I proved the reverse, but I'm stuck on $(\Rightarrow)$.

Since $E$ is measurable all of the $A_k's$ are bounded and measurable. $f$ is integrable-i.e., its integral is finite, so since $\int_A f=\sup\lbrace\int_A h\,| \rbrace$ each element of the set $\lbrace \int_A h\rbrace$ where $0\le h\le f$ and $h$ is simple is finite. How do I use this to describe the summability of the set $\lbrace \mu(A_k) \rbrace $?

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    Have you tried the contrapositive? It might be easier to assume that $\{\mu(A_k)\}$ isn't summable, then construct a simple function $0\leq h\leq f$ that has infinite integral.2012-11-12

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Let $F$ be the set where $f$ is finite.

Suppose $x \in F$. Then for some $n$, we have $n\leq f(x) < n+1$. Hence $x \in A_1 \cap \cdots \cap A_n$, but $x \notin A_k$ for $k > n$. Hence we have $n=1_{A_1}(x)+\cdots + 1_{A_n}(x) = \sum_{k=1}^\infty 1_{A_k}(x) \leq f(x)$.

If $x \notin F$, we trivially have $\sum_{k=1}^\infty 1_{A_k}(x) \leq f(x) = \infty$.

Then by the monotone convergence theorem we have $\sum_{k=1}^\infty \mu A_k = \sum_{k=1}^\infty \int 1_{A_k} d \mu = \int \sum_{k=1}^\infty 1_{A_k} d\mu \leq \int f d\mu < \infty$.

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    I'm not sure what you mean. We can define the measurable set $A_\infty=\{x|f(x) = \infty\}$. It may be empty, but in general, integrability of $f$ just implies that $\mu A_\infty = 0$.2012-11-12