1
$\begingroup$

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???

  • 0
    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..2012-10-01

1 Answers 1

1

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.