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First, I give the definition from Folland,

Definition: Let $(X, \mathcal{M}, \mu)$ be a measure space. If for each $E \in \mathcal{M}$ with $\mu(E) = \infty$, there exists $F \in \mathcal{M}$ with $F \subset E$ and $0 < \mu(F) < \infty$, $\mu$ is called semifinite.

Now problem:

Let X be any nonempty set, $\mathcal{M} = \mathcal{P}(X)$, and $f$ any function from $X$ to $[0, \infty]$. Then $f$ determines a measure $\mu$ on $\mathcal{M}$ by the formula $\mu(E) = \sum_{x \in E} f(x)$.

In the later paragraph, Folland states that,

The reader may verify that $\mu$ is semifinite iff $f(x) < \infty$ for every $x \in X$.

I ask that, how to verify that? Thanks.

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If $f(x)=+\infty$ for some $x$, then $\{x\}\in\cal M$ and is of infinite measure. We cannot find a subset of finite positive measure.

If $f(x)<\infty$ for all $x\in X$, and $\mu(E)=+\infty$, let $x\in E$ such that $\mu(\{x\})>0$. Then $\mu(\{x\})$ is finite, and $\{x\}\subset E$ is measurable.

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OK, but I have one more question. We say $f(x) < \infty$ for all $x \in X$, then how can we say that $\mu(E) = \sum_{x \in E} f(x) = \infty$ where $E \in \mathcal{M}$? Is it related to uncountable sums? Then, how can we verify that this $E$ has a subset $F$ which gives a finite measure? –  Deniz Nov 10 '12 at 14:07
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$\mu(E)$ infinite means that either uncountably many elements of $E$ have positive values of $f$, or countably many and in this case the series is divergent. –  Davide Giraudo Nov 10 '12 at 14:13
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