Suppose $\{f_n\}$ is a sequence of measurable functions on $[0,1]$. For any $x \in [0,1]$, suppose that $\exists$ $N$ s.t. $f_n(x)=0$ for all $n\ge N$. Define function $h$ as the smallest index $n$ for which $f_n(x)=0$, i.e. $h(x) = inf\{n:f_n(x)=0\}$.
If I want to show that h is measurable function, what should I do?
Actually, I've tried a lot. I thought my definition of $g_n := \chi_{\{0\}}(f_n(x))$ is good because it used the condition that $\{f_n\}$ is measurable, but not useful to prove it. Or definition? But I still don't know how to show the inverse is a $\sigma$-algebra.
We have $$h^{-1} (\{k\}) =\{x\in [0,1] :\inf\{n: f_n (x) =0\} =k\}=f_k^{-1} (\{0\} ) \setminus \left(\bigcup_{1\leq j\leq k-1} f_j^{-1} (\{0\} )\right)$$ and since the functions $f_j$ are mesurable therefore $h^{-1} (\{k\}) $ is measurable for any $k\in\mathbb{N}.$ Now since $h$ attains only natural values the above implies measurability of $h$.