measure theory: how does a function f in L1 induces a measure? In a measure space (X,$\mu$), let f $\in L^1$ and f$\ge$0. for every measurable set E let $\mu_f$=$\int_E$f = $\int_X$f$\chi_E$
1) show $\mu_f$ is a measure
do I have to show that f induces a measure on E of X? I'm no sure how to approach this problem. I have been thinking on using simple functions to approximate f but that's to show how $\mu_f$ is defined not $\mu_f$ as a measure. I haven't proved anything like this before so some extra details would be apreciated.
2) show if $\mu(E)$=0, then $\mu_f$=0
so I guess an f-induced measure here is dependent on the measurable space. I don't even know how $\mu$ is defined. Is E supposed to be the empty set in this case? 
3) provide an examples where $\mu_f$=0 but $\mu$>0.
I was thinking the only way this is possible is if the function f sends everything in E to 0, but E is not empty so the induced metric doesn't measure everything.
I greatly appreciate your help, I'm trying to learn this well, so that I can do better dealing with dense functions in L1 and integrable functions in general. 
 A: 1) So $\mu_f(E)= \int \chi_E f d\mu$ where $\chi_E$ is the characteristic function of $E$ (for $E$ a $\mu$-measurable set). So what are the axioms you need to check for $\mu_f$ to be a measure? First there would be $\mu_f(\emptyset)=0$, but this is easy right? (Do you see why?) What about the second axiom? Let $\{E_n\}$ be a pairwise disjoint, measurable collection of sets. We need to verify that $$\mu_f(\bigcup E_n) = \sum \mu_f(E_n)$$ right? Since the $E_n$ are pairwise disjoint we have $\chi_{\bigcup E_n} = \sum \chi_{E_n}$ (check this!). But then $$\mu_f(\bigcup E_n) = \int f \sum \chi_{E_n} d\mu = \sum \int f \chi_{E_n} d\mu$$  (Why was I able to interchange the series and the integral?) Thus $\mu_f$ is a measure.
2) If $E$ is $\mu$-null, i.e. $\mu(E)=0$, what can you say about $f \chi_E$?
3) There are many ways to go about this last question. The trivial one is to just let $f =0$. Yet we can do better with not too much technicalities, just let $\mu$ be the Lebesgue measure on the Borel sigma algebra of $\mathbb{R}$ and take any $\mu$-nullset, let's say $N$ is such a set, and define $f= \chi_N$. What is $\mu_f$ then?
A: Hint: Suppose $E_1,E_2,\dots $ are pairwise disjoint and measurable. Then
$$\chi_{E_1\cup E_2 \cup \dots} = \sum_n\chi_{E_n}.$$
