# Proving that the pointwise limit function of sequence of measurable functions is measurable without limsup or liminf

Let $\{f_n\}$ be a sequence of Lebesgue measurable real valued functions defined on some Lebesgue measurable set E. Suppose the sequence converges pointwise on $E$ to some function $f$. Prove that $f$ is Lebesgue measurable.

Most of the proofs I have seen use the limit superior and limit inferior of $f$. I would like an alternative proof that does not use this. Are there any?

Thanks.

Such a proof can be found here.

Let me translate.

Consider $U$ an open subset of $\mathbb R$. Let $U_n=\{x\in U, \;\; d(x, \mathbb R\setminus U)>\frac 1n\}$ and $F_n=\{x\in U , \;\; d(x, \mathbb R\setminus U)\geq\frac 1n\}$.

Note that: each $U_n$ is open, each $F_n$ is closed, $\cup_n U_n = U$ and $\cup_n F_n=F$.

The following inclusions hold:

\begin{align} f^{-1}(U) &= \cup_n f^{-1}(U_n) \\ &= \cup_n \{x, \;\; \lim_k f_k(x)\in U_n\}\\ &\subset \cup_n \{x, \;\; \exists K, \forall k\geq K, f_k(x)\in U_n\} \\ &\subset \cup_n \cup_K \cap_{k\geq K} \;f_k^{-1}(U_n)\\ \end{align}

Note that $\cup_n \cup_K \cap_{k\geq K} \;f_k^{-1}(U_n)$ is a member of the sigma-algebra you defined on $E$.

Since $U_n\subset F_n$, $$\cup_n \cup_K \cap_{k\geq K} \;f_k^{-1}(U_n) \subset \cup_n \cup_K \cap_{k\geq K} \;f_k^{-1}(F_n)$$

Note that for any $n$, $$\cup_K \cap_{k\geq K} \;f_k^{-1}(F_n) = \{x, \exists K, \forall k\geq K, f_k^{-1}(x)\in F_n\}\subset \{x, f(x)\in F_n\}= f^{-1}(F_n)$$

Therefore,

\begin{align} \cup_n \cup_K \cap_{k\geq K} \;f_k^{-1}(U_n) &\subset \cup_n \cup_K \cap_{k\geq K} \;f_k^{-1}(F_n)\\ &\subset \cup_n f^{-1}(F_n)\\ &= f^{-1}(\cup_n F_n)\\ &= f^{-1}(U) \end{align}

Conclusion: $$f^{-1}(U) = \cup_n \cup_K \cap_{k\geq K} \;f_k^{-1}(U_n) \in \Sigma$$