Prove that the following set has full measure (i.e. has probability one): Let $(X,\mathscr{A},\mu)$ be a probability space. Let $f\in L^{1}(X,\mathscr{A},\mu)$ be such that $f(X)\subset\mathbb{N}$. We construct a new probability space as follows:


*

*Set: $X_{f}:=\{(x,k) \ | \ x\in X, \ k\in\mathbb{N}, \ 1\leq k \leq f(x)\}\subset X\times\mathbb{N}$.

*Sigma algebra: $\mathscr{A}_{f}$ is generated by sets of the form $$(A\times\{n\})\cap X_{f}=\{(x,n) \ | \ x\in A, \ f(x)\geq n\},$$ where $A\in\mathscr{A}$ and $n\in\mathbb{N}$.

*Probability measure: Define $\mu_{f}$ on a generator $(A\times\{n\})\cap X_{f}$ by $$\mu_{f}((A\times\{n\})\cap X_{f}):=\frac{1}{\int_{X}f \ \text{d}\mu}\mu(A)$$ and extend $\mu_{f}$ to all of $\mathscr{A}_{f}$.


Now I have two questions:


*

*I dont really see why $\mu_{f}(X_{f})=1$ and,

*If $X^{\circ}\subset X$ has full $\mu$-measure (i.e. $\mu(X^{\circ})=1$), then how do I prove that the set $$\{(x,k) \ | \ x\in X^{\circ}, \ k\in\mathbb{N}, \ 1\leq k\leq f(x)\}$$ has full $\mu_{f}$-measure?

 A: Let $X_{k}=\{x\in X: f(x)\geq k\}$, then 
\begin{align*}
X_{f}=\bigcup_{k\in\mathbb{N}}(X_{k}\times\{k\})\cap X_{f},
\end{align*}
and hence
\begin{align*}
\mu_{f}(X_{f})&=\sum_{k=1}^{\infty}\mu_{f}((X_{k}\times\{k\})\cap X_{f})\\
&=\left(\int_{X}fd\mu\right)^{-1}\sum_{k\in\mathbb{N}}\mu(X_{k})\\
&=\left(\int_{X}fd\mu\right)^{-1}\sum_{k\in\mathbb{N}}\mu(\{x\in X:f(x)\geq k\})\\
&=\left(\int_{X}fd\mu\right)^{-1}\sum_{k\in\mathbb{N}}\sum_{l\geq k}\mu(\{x\in X:f(x)=l\})\\
&=\left(\int_{X}fd\mu\right)^{-1}\sum_{l\in\mathbb{N}}\sum_{k\leq l}\mu(\{x\in X:f(x)=l\})\\
&=\left(\int_{X}fd\mu\right)^{-1}\sum_{l\in\mathbb{N}}l\mu(\{x\in X:f(x)=l\})\\
&=\left(\int_{X}fd\mu\right)^{-1}\sum_{l\in\mathbb{N}}\int_{X}f1_{\{x\in X:f(x)=l\}}d\mu\\
&=\left(\int_{X}fd\mu\right)^{-1}\int_{X}fd\mu\\
&=1.
\end{align*}
For the second question, proceed the above procedures replacing $X$ by $X^{\circ}$, one has
\begin{align*}
\mu_{f}(\text{the set in question})=\left(\int_{X}fd\mu\right)^{-1}\int_{X^{\circ}}fd\mu.
\end{align*}
But now $X$ and $X^{\circ}$ just differ a set of measure zero, so $\displaystyle\int_{X^{\circ}}fd\mu=\int_{X}fd\mu$.
