# Lebesgue integral over subsets

Let $v(E)=\int_{E} f d\mu=\mu (1_{E}f)$. Take $f \in \Sigma^{+}$ a non-negative measurable function and h a measurable function, $h\in \Sigma$. I now need to show that $h \in L^{1}(S,\Sigma,v)$ if and only if $hf \in L^{1}(S,\Sigma, \mu)$. If this holds then $\int h d\nu=\int hf d\mu$. Can anyone help me with this???

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It is often instructive in such problems to return to the definitions of the concepts at hand. In this case, recall that a function being in $L^1$ can be defined (and often is defined) by saying that the function can be approximated by a Cauchy sequence of step functions, and the $L^1$-norm of the function is equal to the limit of the $L^1$ norms of the approximating step functions. In this case, why don't you try using that definition, since once you have step functions, we can more easily pass between measures $\mu$ and $\nu$ since on each step, the function is constant. –  Christopher A. Wong Oct 1 '12 at 7:58
The only thing we've had so far is that a function f is lebesgue measurable if $\int f^{+} d\mu<\infty$ and $\int f^{-}d\mu<\infty$. If a function $g$ is lebesgue integrable and non-negative then we can write the integral of $g$ as the supremum of the integral of a simple function if the simple function smaller or equal than g. –  Math Girl Oct 1 '12 at 8:24
OK, good, so to verify that $h$ is in $L^1(\nu)$, we want to verify that $h^{+}$ and $h^{-}$ are in $L^1(\nu)$. Let's try to do that by now approximating those two things by simple functions, and then use the formula you have written for converting $\nu$ into $\mu$. –  Christopher A. Wong Oct 1 '12 at 8:31
I've tried this.. As $h^{+}$ and $h^{-}$ are now non-negative functions we can write their lebesgue integral as the supremum of simple functions like $$\int h^{+} d\nu=sup\{\int k d\nu;k\leq h^{+},k \in S^{+} \}$$ As a simple function can be written as $k=\sum_{i}\alpha_{i}1_{A_{i}}$, the integral of a simple function can be written as $\int k d\nu=\sum_{i}\alpha_{i}\nu(A_{i})$. We now have that $\nu(A_{i})=\int_{A_{i}} f d\mu$. This is as far as I've gotten by myself. I don't know how to precede from here.. –  Math Girl Oct 1 '12 at 16:46

Proof this first for the indicator function, then for simple functions. After you prove this for these functions then for every $h^{+}$ you can find a sequence of increasing simple functions $h_{n}^{+}$ with limit $h^{+}$ and then use the monotone convergence theorem.