# Lebesgue Integrabilityon relative measure space

This shouldn't be too difficult but I just can't get started. There are quite a few similar problems on my problem set so I am hoping to get some idea here on this one, and try solving the others afterwards.

Let $(X,M,\mu)$ be a positive measure space. Let $E\in M$, and $(E,M_{E},\lambda)$ , where $\lambda(A)=\mu(A\cap E)$, be the relative measure space. Now if $f$ is Lebesgue integrable with respect to $(X,M,\mu)$, show that $f|_{E}$ is Lebesgue integrable with respect to $(E,M_{E},\lambda)$, and that $\int_{E} f|_{E} d\lambda = \int_{X} f \chi_{E} d\mu$.

I first showed $f|_{E}$ is measurable with respect to $(E,M_{E},\lambda)$: since $f$ is measurable, $\{x:f(x)>a\}\in M, \forall a \in R$, and taking the intersection with E on both sides shows that $f|_{E}$ is measurable.

Then I tried various ways to proceed (mostly through definitions of Lebesgue integrability, such as separating $f$ into positive and negative parts, then approximating them with simple functions), but all attempts failed. Is there a "standard" way to show something is Lebesgue integrable? And how do I link two different measure spaces together? Any ideas would be greatly appreciated!

I think I may have figured this out. See if these steps are right: $\int_{X} f \chi_{E} d\mu = \int_{E} f |_{E} d\mu$ , since $f |_{E} = f \chi_{E}$ for all x in E. But then $\int_{E} f |_{E} d\mu = \int_{E} f |^{+}_{E} d\mu - \int_{E} f |^{-}_{E} d\mu$, where the +/- denotes the positive and the absolute value of the negative part of the function. And again by definition, $\int_{E} f |^{+}_{E} d\mu - \int_{E} f |^{-}_{E}d\mu = sup\{\int_{E} s^{+}_{E} d\mu\} - sup\{\int_{E} s^{-}_{E}d\mu\}$, where the $s$'s are simple functions approximating the $f$'s. From there we apply the definition of Lebesgue integral of simple functions, where we finally have the chance to change the measure from $\mu$ to $\lambda$, and then we are basically finished.
All you need is for the positive and negative parts to have finite integral. Since $\chi_E$ is $0$-$1$ valued you have $$\int_E f^+|_E \, d\mu = \int_X f^+ \chi_E \, d\mu \le \int_X f^+ \, d\mu < \infty$$ and similarly $$\int_E f^-|_E \, d\mu = \int_X f^- \chi_E \, d\mu \le \int_X f^- \, d\mu < \infty.$$