Definition 1:

Say $X = \bigcup_{n=1}^{\infty} E_n $ where $E_n \in \mathcal{M}$ and $\mu( E_n ) < \infty $ for all $n$, we call $\mu$ $\sigma$-finite. More generally, if $E = \bigcup^{\infty} E_n $ where $E_n \in \mathcal{M}$ for all $n$ and $\mu(E_n) < \infty $ for all $n$ then $E$ is said to be $\sigma-$finite for $\mu$.

Definition 2:

If for each $E \in \mathcal{M}$ with $\mu(E) = \infty$, there exists $F \in \mathcal{M}$ with $F \subset E $ and $0 < \mu(F) < \infty$, then we cal $\mu$ semifinite.


Every $\sigma-$finite measure is semifinite.


Let $\mu$ be a $\sigma-$finite measure on $X$. Take $E \in \mathcal{M}$ arbitrary with $\mu(E) = \infty $. Write

$$ E = \bigcup^{\infty} E_n \; \; \; \; E_n \in \mathcal{M} \; \; forall \; \; n $$

(Here is where I am not sure I am doing the problem correctly. Can I assume that I can write $E$ in such a form? )

Next, there is some $k$ such that $E_k \subset E $. Hence by monotonicity,

$$ 0 \leq \mu(E_k) < \mu(E) = \infty $$

  • $\begingroup$ You can write $E$ in such a form by taking $E_n = X_n \cap E$, where $X =\bigcup X_n$ and each $X_n$ is of finite measure. In your last equation, you have $0\leq ...$ as the first step. But you need $0<...$. Also, you do not have $E_n \subset E$ for some $k$,but in fact for all $k$ (why? and how does that help?). $\endgroup$
    – PhoemueX
    Nov 25, 2014 at 22:50
  • $\begingroup$ Ok, I understand $E_n \subset E$ for all $n$. but we only need one of these, say $E_k$. if $\mu(E_k) = 0 $, then $\mu(E) = 0 $ since this holds for all $n$. contradiction. Am I on the right track? $\endgroup$
    – ILoveMath
    Nov 25, 2014 at 23:05
  • $\begingroup$ Yes, if all $E_n$ have measure zero, then so has $E$, contradiction. Hence $0 < \mu(E_n) < \infty$ for some $n$. $\endgroup$
    – PhoemueX
    Nov 26, 2014 at 9:20

1 Answer 1


Just prove it by the contradiction. Suppose that $\mu$ is not semifinite, so there is a set $E\in\mathcal M$ such that $ \mu(E)=+\infty $ and for any $F\in\mathcal M$ with $F\subset E$, we have $\mu(F)\in\{0,\infty\}$.

Suppose that $\mu$ is $\sigma$-finite, then there are $E_1, E_2, ..., E_n,... \in \mathcal M$ such that $X=\bigcup_{n=1}^\infty E_n$ and $\mu(E_n)<\infty$ for any $n\in\mathbb N$. Then we have $$ E=\bigcup_{n=1}^\infty (E\cap E_n)\ \Rightarrow\ \mu(E)\leq \sum_{n=1}^\infty \mu(E\cap E_n) $$ and since $\mu(E\cap E_n)\leq\mu(E_n)<\infty$ and $\mu(E\cap E_n)\in\{0,\infty\}$, $\mu(E\cap E_n)=0$ and consequently, $\mu(E)=0$, a contradiction.


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