# Every $\sigma-$finite measure is semifinite. $(X, \mathcal{M}, \mu)$ is a measure space.

### 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.

### Problem:

Every $$\sigma-$$finite measure is semifinite.

### Attempt

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,

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

• 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?). Nov 25, 2014 at 22:50
• 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? Nov 25, 2014 at 23:05
• Yes, if all $E_n$ have measure zero, then so has $E$, contradiction. Hence $0 < \mu(E_n) < \infty$ for some $n$. Nov 26, 2014 at 9:20

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.