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If $\mu$ is semifinite measure and $\mu(E) = \infty$, for any $C > 0$ there exists $F\subset E$ with $C < \mu(F) < \infty$.

Attempted proof - Suppose $E\in M$ with $\mu(E) = \infty$ then there exists $F\in M$ with $F\subset E$ and $0 < \mu(F) < \infty$, $\mu$ is called semifinite. This is the definition from Folland. I am guessing that we have to set $C$ equal to something where the definition still holds true but I am not sure how to do this. Any suggestions is greatly appreciated.

Wolfy
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2 Answers2

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Every set with infinite measure contains a set with finite positive measure.

Define $T = \sup\{ \mu(F) \mid F \subset E,\ \mu(F) < \infty,\ \text{$F$ measurable}\}$.

Clearly $T > 0$ and in fact $T = \infty$. If to the contrary $T < \infty$ then for every $k \ge 1$ there exists a measurable set $F_k \subset E$ with the property that $T - \frac 1k < \mu(F_k) \le T$.

Define $G_k = F_1 \cup \ldots \cup F_k$. Then each $G_k$ is a measurable subset of $E$ with $\mu(G_k) < \infty$ so in fact $\mu(G_k) \le T$.

Define $G = \cup G_k$. Then $G \subset E$, and by the continuity of the measure $\mu(G) \le T$. Since $F_k \subset G_k \subset G$ we obtain $$T - \frac 1k < \mu(G) \le T$$ for all $k$, so that $\mu(G) = T$.

Now, $\mu(E \setminus G) = \infty$ so that $E \setminus G$ contains a measurable set $H$ with positive finite measure. Then $G \cup H \subset E$, and $$\mu(G \cup H) = T + \mu(H) > T$$ contrary to the definition of $T$. Thus $T = \infty$.

At this point the argument is complete, but to recap if $C > 0$ then $C < \sup\{\mu(F) \mid F \subset E,\ \mu(F) < \infty,\ \text{$F$ measurable}\}$ so in particular there exists a measurable subset $F$ of $E$ satisfying $C < \mu(F) < \infty$

Umberto P.
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Let $$S=\{x\,|\,\exists F\subset E: x\le\mu(F)<\infty\}$$ and $s=\sup S$. Since $\mu$ is semifinite, $S$ is non-empty and $s>0$. We want to show that $s=\infty$. This can be done by contradiction, that is, suppose that $s<\infty$. Then it is possible to show

  1. $s\in S$: For each $n$, we can find $F_n\subset E$ so that $s-\frac1n\le \mu(F)<\infty$. Consider two cases:

    • $\bigcup F_n$ has finite measure.
    • $\bigcup F_n$ has infinite measure.
  2. Since $s\in S$, we can find $F\subset E$ so that $s\le \mu(F)<\infty$. Here, consider two cases:
    • $\mu(F)=s$
    • $\mu(F)>s$
Quang Hoang
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