I have come across this question in which the author quoted the following content from a lecture note. The author only provided a screenshot and already exited MSE. I'm very interested in these notes.
Do you know where is the source of these notes?
Thank you so much for your reference!
Consider a measure space $(\Omega, \Sigma, \mu)$, and a Hausdorff locally convex space $X$. A function $f: \Omega \rightarrow X$ is said to be measurable if $f^{-1}(G) \in \Sigma$ for every open set $G \subset X$. The function $f$ is said to be weakly measurable if $\varphi \circ f$ is measurable for every $\varphi \in X^{*}$; clearly measurability implies weak measurability. A measurable function which only takes finitely many values is called a simple function. The function $f$ is said to be strongly measurable if there exist simple functions $f_{n}: \Omega \rightarrow X$ such that $f(\omega)=\lim _{n \rightarrow \infty} f_{n}(\omega)$ for $\mu$-almost every $\omega \in \Omega$. It is clear that a limit of measurable functions is measurable, and therefore strongly measurable functions are measurable. It is also clear that, except for a set of measure zero, the values of a strongly measurable function belong to a separable subspace of $X$. The converse also holds if $\mu$ is $\sigma$-finite and $X$ is metrizable; this follows from the following result of Pettis.
THEOREM 3.6.1. If $X$ is a separable, metrizable locally convex space, $(\Omega, \Sigma, \mu)$ is a $\sigma$-finite measure space, and $f: \Omega \rightarrow X$ is weakly measurable, then $f$ is strongly measurable.
PROOF. We can restrict ourselves to the case of finite measures. We prove first that $f$ is measurable. To do this it suffices to show that $f^{-1}(x+V) \in \Sigma$ whenever $x \in X$ and $V$ is a neighborhood of zero and a barrel. Replacing $f$ by $f-x$, we may also assume that $x=0$. Chose points $y_{n} \in X \backslash V$, and open convex neighborhoods such that $\Re \varphi_{n}\left(y_{n}+v_{n}\right)>\Re \varphi_{n}(v)$ whenever $v_{n} \in V_{n}$ and $v \in V$. Then the set $f^{-1}(V)=\bigcap_{n=1}^{\infty}\left\{\omega \in \Omega: \Re f(\omega) \leq \sup _{v \in V}\left|\varphi_{n}(v)\right|\right\}$ is a countable intersection of sets in $\Sigma$. To show strong measurability, fix a neighborhood base at zero $V_{1}, V_{2}, \ldots$, and note that for each $n$ there exist points $x_{k}^{n}$ such that $X=\bigcup_{k=1}^{\infty}\left(x_{k}^{n}+V_{n}\right)$. For each $n$, the sets $\sigma_{k}^{n}=\left\{\omega: f(\omega) \in\left(x_{k}^{n}+V_{n}\right) \backslash \bigcup_{j<k}\left(x_{j}^{n}+V_{n}\right)\right\}$ form a partition of $\Omega$, and we can choose $N_{n}$ so that $\mu\left(\bigcup_{k>N_{n}} \sigma_{k}^{n}\right)<2^{-n}$. The simple functions $f_{n}=\sum_{k=1}^{N_{n}} \chi_{\sigma_{k}^{n}} x_{k}^{n}$ converge almost everywhere to $f$.