Any help with this question?

Let $X$ be an nonempty set and $f\colon X\to[0,+\infty]$ a function. Defines $$\sum_{x\in E} f(x):=\sup\left\lbrace\sum_{x\in F}f(x);F\subset E\ is\ finite\right\rbrace.$$ Consider the $\sigma$-algebra $\mathcal{F}=\mathcal{P}(X)$ and defines $$\mu(E):=\sum_{x\in E}f(x).$$ Show that:

(a) $\mu$ is a measure.

(b) $\mu$ is semifinite $\Longleftrightarrow$ $f(x)<+\infty,\forall x\in X$.

(c) $\mu$ is $\sigma$-finite $\Longleftrightarrow$ $\mu$ is semifinite and $\{x\in X;f(x)>0\}$ is countable.

I was think in the case to show that $\mu(\emptyset)=0$ and get already stacked. So i can't move on... Thanks for any help!

To me, doesn't make sense $$\mu(\emptyset)=\sup\left\lbrace\sum_{x\in F}f(x);F\subset\emptyset\ is\ finite\right\rbrace,$$ because in this case $F=\emptyset$, so, how i conclude that the $sup$ of this set is $0$? what is the sum of $f(x)$ with $x\in\emptyset$?

  • 3
    $\begingroup$ I think you probably need to define the empty sum to be $0$. $\endgroup$ – David Kraemer Apr 17 at 11:58
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    $\begingroup$ At least you need to define $\sum_{x \in \varnothing} f(x)$ somehow, or else your definition is useless. $\endgroup$ – GEdgar Apr 17 at 12:37
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    $\begingroup$ The best way would be to define it to be zero. But you could argue that by vacuity it should be zero, since there are no elements to sum. $\endgroup$ – AspiringMathematician Apr 17 at 12:46

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