I have a question about a nonnegative Lebesgue measurable function $f:[0,1] \rightarrow \mathbb{R} \cup \{ \infty \} $.
The question asks me to show that $$f= \sum_{n=0}^ \infty f \chi_{f^{-1} ([n, n+1) )} .$$
But unless I’m missing something this is only true for when $f$ is finite valued. If $f(x)= \infty $ then $\chi_{f^{-1} ([n, n+1))}(x)=0 $ for all $n$ so it can’t possibly hold for this $x$.
You are right. Let $x\in[0,1]$. If we have $f(x)<\infty$, then we have $f(x)\chi_{f^{-1}([n,n+1)}(x)=0$ for all $n\neq\lfloor f(x)\rfloor$ since then $f(x)\not\in[n,n+1)$ and hence $x\not\in f^{-1}([n,n+1)$, while for $n=\lfloor f(x)\rfloor$ we have $n\ge 0$ because $f$ is non-negative, and $n<\infty$ since $f(x)<\infty$, so $f(x)\chi_{f^{-1}([n,n+1)}(x)=f(x)$ because $f(x)\in[n,n+1)$ and thus $x\in f^{-1}([n,n+1))$. If we have $f(x)=\infty$, then we have $f(x)\not\in[n,n+1)$ for all $n\ge 0$ and thereby $x\not\in f^{-1}([n,n+1))$, which shows that $f(x)\chi_{f^{-1}([n,n+1)}(x)=0$. In total, this gives $f\chi_{f^{-1}([0,\infty))}=f\chi_{f^{-1}([n,n+1))}$.