Htykuut

Let $(X, mathcal{A},mu)$ be a measure space and $h:X to mathbb{R}, h geq0$ a measurable function.

Define a map $mu_h:mathcal{A} to mathbb{R}, mu_h(E):= int_E h dmu$ for $E in mathcal{A}$.

How can it be shown that if $mu(E)=0 Rightarrow mu_h(E)=0$?

To prove this implication I tried:

$int_E h dmu=int_E h_+ dmu-int_E h_- dmu$

So $int_E h_+ dmu=int h_+1_E dmu=$sup$int e dmu$, with $e$ is a simple function.

Here I don't know how to continue to show this implication. Or is there another way to prove it?

By definition $mu_h(E)=int hmathbf1_E dmu=sup{int sdmumid stext{ is a measurable simple function with } 0leq sleq hmathbf1_E}$

So it is enough to prove that $mu(E)=0$ implies that $int sdmu=0$ for every measurable simple function $s$ that satisfies $0leq sleq hmathbf1_E$.

Such function $s$ can be written as a finite sum $sum_{i=1}^n a_imathbf1_{A_i}$ where the $a_i$ are nonnegative and the $A_i$ are measurable subsets of $E$.

So we have $int sdmu=sum_{i=1}^n a_imu(A_i)=0$ because $0leqmu(A_i)leqmu(E)=0$ for every $iin{1,dots,n}$.

The function $g=h1_E$ is approximated below by $f_n(x)=sum_{i=0}^{n2^n-1}i/(n2^n)1_{{i/(n2^n)leq g(x)<(i+1)/(n2^n)}}$ and $int f_ndmu=0$ for all $n$.

So, $mu_h(E)=int gdmu=sup{int sdmu|0leq sleq g, s text{ is a simple function}}=lim int f_ndmu=0$.