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Let $(X,\mathcal{E},\mu)$ be a measure space.

Definition: The semi-finite version of $\mu$ on $(X,\mathcal{E})$ is denoted $\mu_{\rm sf}$ and defined as $$ \mu_{sf}(E) = \sup\left\{\mu(F);\; F\subseteq E,\,F\in \mathcal{E}\; \text{and}\; \mu(F)<\infty\right\}, \; \forall\,E \in \mathcal{E} . $$

I want to understand the proof of the following lemma.

Lemma: Suppose $E \in \mathcal{E}$ with $\mu_{sf}(E) < \infty$. Then there exists a measurable $B \subseteq E$ with $\mu(B) = \mu_{sf}(E)$.

Proof:By definition of $\mu_{sf}$, for each $k>0$ we may find a measurable $A_k \subseteq E$ with $\mu(A_k)<\infty$ and $\mu_{sf}(E) - \frac{1}{k}\leq \mu(A_k)$. Set $B_n = A_1 \cup \dots \cup A_n$. Then $B_n \subseteq E$ and $\mu(B_n) \ge \mu_{sf}(E) - \frac{1}{n}$. Moreover, since $\mu(B_n) < \infty$, we have $\mu(B_n) \le \mu_{sf}(E)$ by the definition of $\mu_{sf}$. Now set $B = \bigcup_{n=1}^\infty B_n$. Continuity from below shows $\mu(B) \le \mu_{sf}(E)$ and monotonicity shows $\mu(B) \ge \mu_{sf}(E)$. So $B$ is as desired.

I don't understand the following two facts:

  • $(1)$ Why continuity from below shows that $\mu(B) \le \mu_{sf}(E)$?

  • $(2)$ Why monotonicity shows that $\mu(B) \ge \mu_{sf}(E)$?

Student
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1 Answers1

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Since $B_1 \subset B_2 \subset ...$ continuity from below implies that $\mu(B) = \lim_n \mu(B_n)$. Since $\mu(B_n) \leq \mu_{sf}(E)$ holds for all $n$, this relation holds in the limit too. Hence, $\mu(B) \leq \mu_{sf}(E).$

Since $B_n \subset B$ for all $n$, monotonicity implies that $\mu(B) \geq \mu(B_n) \geq \mu_{sf}(E) - 1/n$ holds for all $n$. Letting $n \to \infty$ on the right-hand side shows that $\mu(B) \geq \mu_{sf}(E)$.

aduh
  • 8,662
  • Thank you very much for your answer. But could you please explain me this fact:''Since $B_1 \subset B_2 \subset ...$ continuity from below implies that $\mu(B) = \lim_n \mu(B_n)$.''? – Student Apr 01 '18 at 09:27
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    @Student What I wrote is nothing more than the definition of continuity from below. In particular, continuity from below means: For any measure space and any increasing sequence $(B_n)$ of measurable sets, the union of the sequence is measurable and has measure equal to the limit of the measures of $B_n$. This property has been discussed many times on this site (here, for example) and is covered in all the main textbooks on measure theory. – aduh Apr 01 '18 at 12:13