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Let $X\neq \emptyset$ and $f:X \rightarrow [0, \infty]$ not identical infinity. Set $$ \sum_{x \in X} f(x)= \sup \left\{ \sum_{x \in F}f(x), F \subseteq X, F \mbox{ finite} \right\}.$$ $(i)$ Show that $\mu(E)= \sum_{x \in E} f(x)$ is a measure on $(X, P(X))$;

$(ii)$If $f(x) < \infty \> \forall x \in X$ and the set $\lbrace{ x \in X: f(x)>0 \rbrace}$ is at most countable show that $\mu$ is $\sigma$-finite.

My solution use this argument:

$(i)$ Let $E=\cup_n E_n $ with $E_n$ pairwise disjoint; If $x$ doesn't belong to $E$ then $x$ doesn't belong to any of the $E_n$ and this means $$ \mu(E)=0= \sum_{x \in \cup_n E_n} f(x)$$ Now let's consider that the set $E$ contains just one element $\overline{x}$. If $\overline{x} \in E$ then it exists a unique $\overline{n}$ such that $\overline{x} \in E_{\overline{n}}$. Computing the measure of $E$ leads to: $$ \mu(E)=f(\overline{x})=\mu(\cup_n E_n)=\mu(E_{\overline{n}})+ \sum_{n\neq \overline{n}} f(x)=\mu(E_{\overline{n}})+\sum_{n\neq \overline{n}} \mu(E_n)$$ that should hold because $\sum_{n\neq \overline{n}} \mu(E_n)=0$

$(ii)$ Being that $A=\lbrace{ x \in X : f(x)>0 \rbrace}$ is at most countable we can write $$ A=\cup_n \lbrace{x_n\rbrace}$$ and $\mu(\lbrace{x_n\rbrace})=f(x_n) <\infty$ from the hypothesis.

It's my solution acceptable? I didn't use the definition given with the $\sup$.

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You are using the sum as the standard sum, however in your case it is just notation for $\sup \lbrace{ \sum_{x \in F}f(x), F \subseteq X, F \text{finite} \rbrace}$. The proof is similar to showing that inner measure is a measure. You also need to show $\mu(\varnothing)=0$. –  user10444 May 4 at 13:38
    
oh ok... I did see that there is something similar with the definition of the outer measure but I didn't get into deep. The outer measure is a well defined measure just on the Lebegue $\sigma$-algebra...in this case the measure is defined for all sets in $P(X)$. Does this follow from the fact that this one gives a not null measure also to singletons? –  user73793 May 4 at 13:47
    
Well you can define Lebesgue outer measure to any set but we restrict it to the Lebesgue $\sigma$-algebra so we get $\lambda(E)=\lambda^*(E)$ for $E$ in the Lebesgue algebra, and it maintains $\sigma$-additivity. –  user10444 May 4 at 13:51
    
Uhm... From what I have been studying these last days the Lebesgue outer measure is countable addictive (so it's a measure) just on a subset of $P(\mathbb{R})$. My question was...why in this case this $\mu$ defined on all $P(X)$ is countable addictive on all subsets? –  user73793 May 4 at 13:59
    
I don't think there is a criteria for a measure to be defined on all $\mathcal{P} (X)$, however the reason for this measure to be defined on all $\mathcal{P} (X)$ should appear in in the proof. –  user10444 May 4 at 14:05

1 Answer 1

Here is $(i)$: It is clear that $\mu(E)\ge 0$. $\mu(\varnothing)=\sup \lbrace{ \sum_{x \in F}f(x), F \subseteq \varnothing, F \text{finite} \rbrace}$, then $F$ is necessarily empty thus $\sum_{x \in F}f(x)=0$.

Now let $(E_n)_{n\in \Bbb N}$ be disjoint sets we want to show $\mu(\bigcup_{n\in \Bbb N}E_n)=\sum_{n\in \Bbb N}\mu(E_n)$. We can assume $\mu(E_n)\neq \infty$ since if otherwise we have equality. If $F \subset \bigcup_nE_n$ then by the disjointness of the $E_n$ $F$ can be written as $F=F_1\cup F_2 \ldots F_n$ where $F_i$ is a finite subset of $E_i$ and are disjoint. Now $$\sup \lbrace{ \sum_{x \in F}f(x), F \subseteq \cup E_n, F \text{finite} \rbrace}=\sup \lbrace{ \sum_{x \in F}f(x), F \subseteq \cup E_n, F \text{finite} \rbrace}\ge \sup \lbrace{ \sum_{i=1}^n\sum_{x \in F_i}f(x), F \subseteq E_1 \cup \dots \cup E_n, F \text{finite} \rbrace}=\sum_{i=1}^n\sup \lbrace{ \sum_{x \in F_i}f(x), F_i \subseteq E_i, F_i \text{finite} \rbrace}$$ Now let $n \to \infty$ to get $\mu(\bigcup_{n\in \Bbb N}E_n)\ge\sum_{n\in \Bbb N}\mu(E_n)$.

For the reverse inequality, note that for any $\sum_{i=1}^n\sum_{x \in F_i}f(x) \in \left\{ \sum_{x \in F}f(x), F \subseteq \cup E_n, F \mbox{ finite} \right\}$ we have $\sum_{i=1}^n\sum_{x \in F_i}f(x)< \sum_{n\in \Bbb N}\sup \lbrace{ \sum_{x \in F_i}f(x), F_i \subseteq E_i, F_i \text{finite} \rbrace} $ thus $\mu(\bigcup_{n\in \Bbb N}E_n)=\sum_{n\in \Bbb N}\mu(E_n)$.

For $(ii)$ You need to write $X$ as a countable union of set with finite measure, I can't understand what you wrote. Write $X=A\cup A^c$, you have already found the measure of $A$ now find that of $A^c$.

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