# test for sigma finiteness

A measure $\mu$ on (E,$M$) is $\sigma$-finite if and only if there exists a strictly positive function $f$ in M such that $\mu f<∞$ .

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Maybe this thread gives some inspiration. –  t.b. May 24 '12 at 16:13
If $\mu$ is $\sigma$-finite then there is a sequence of disjoint sets $A_n\in M$ such that $0 < \mu(A_n) < \infty$ and their union is the whole space: $\bigcup_nA_n = E$. Define $$f(x) = \sum_{n\geq 1}\frac{1}{2^n\mu(A_n)}1_{A_n}(x)$$ i.e. $f(x) = \frac{1}{2^n\mu(A_n)}$ for all $x\in A_n$. Clearly, $f$ is strictly positive and $\mu f = 1<\infty$.
Now, let there be a function $f$ which is strictly positive and $\mu f <\infty$. Define $$B_n = \left\{x:\frac1n\leq f(x)\leq n\right\}.$$ Then $B_n$ is an increasing sequence which covers the whole space $E$. Moreover, $$\mu f = \int\limits_E f(x)\mu(dx)\geq \int\limits_{B_n} f(x)\mu(dx)\geq \frac1n \mu(B_n)$$ so $\mu(B_n)\leq n(\mu f)<\infty$. If we define $A_1 = B_1$ and $A_n = B_{n+1}\setminus B_n$ we obtain a countable disjoint cover of $E$ and $\mu(A_n)<\infty$.
@user: swap integration and summation to get $$\mu(f) = \sum_{n\geq 1}\frac{1}{2^n\mu(A_n)} \mu(1_{A_n}(x)) = \sum_{n\geq 1} 2^{-n} = 1.$$ –  t.b. May 24 '12 at 17:12