Strongly measurable function

Remarks

Example: let $X = [0, 1]$ with the Borel σ-algebra and let $f \colon X \to E$ be continuous. The function $s_n$ that equals $f(k/n)$ on $[k/n, (k+1)/n)$ for $0 \le k < n$ and $f(1)$ at $1$ is simple and measurable, and $s_n(x) \to f(x)$ by continuity, so $f$ is strongly measurable.

Non-example: let $X = [0, 1]$ with the σ-algebra $\set{\varnothing, X}$ and $f(x) = x$ with values in $E = \R$. A simple measurable function is then constant, so a pointwise limit of such functions is constant, and $f$ is not strongly measurable.

When $E = \R$ or $E = \C$, a function is strongly measurable if and only if it is measurable. For a measure $\mu$ on $(X, \Sigma)$, a function is called strongly $\mu$-measurable or Bochner measurable when it agrees $\mu$-almost everywhere with a strongly measurable function. The image of a strongly measurable function is a separable subset of $E$, because it lies in the closure of the countable set of values of the $s_n$.