# Show that a measurable function is integrable.

Let $x$ be a nonnegative random variable on a probability space $(X, \Sigma, P)$ such that $\mathbb{E}(x)=1$ and $P\{x>0\}=1$. Define $y:X\to \overline{\mathbb{R}}_+$ by $$y(w)=\frac{1}{x(w)}$$ if $x(w)>0$, and $y(w)=+\infty$ if $x(w)=0$. Define a new probability measure $q:\Sigma\to\mathbb{R}$ as $q(S)=\int_Sx\,dP$ (this is a probability measure, we do not need to check that). Prove that $$\int_X y\,dq<\infty.$$ Can anyone give me some hint on this problem? Many thanks!

Since $q(S) = \int_S x \, dP$, we have $\int 1_S dq = \int 1_S \,x\,dP$, hence it is true for simple functions, and by DCT, true for $P$-integrable $f$, ie $\int f\, d q = \int f\,x \,dP$. This should be enough to answer your question.
In your case, take $f(\omega) = y(\omega)$, then $\int y d q = \int y(\omega) x(\omega) \,dP(\omega) = \int \, dP = 1$.