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

Please help me out with the proof. Thank you.

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Maybe this thread gives some inspiration. – t.b. May 24 '12 at 16:13
Help you with what? You haven't asked a question. – user5137 May 24 '12 at 16:29
up vote 5 down vote accepted

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 $\mu(A_n)<\infty$.

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Thank you fore your quick answer. But I don`t know why μf=1 – user30795 May 24 '12 at 16:45
@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

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