# Measure and function

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 $$x\in\emptyset$$?

• I think you probably need to define the empty sum to be $0$. – David Kraemer Apr 17 at 11:58
• At least you need to define $\sum_{x \in \varnothing} f(x)$ somehow, or else your definition is useless. – GEdgar Apr 17 at 12:37
• 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. – AspiringMathematician Apr 17 at 12:46