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  "body": "\n## Definition\n\nLet $(X, \\Sigma)$ be a [[measurable space|N4KDT83V3S]] and let $E$ be a [[Banach space|0BDTTM24DP]] over $\\R$ or $\\C$.\nA function $s \\colon X \\to E$ is a *simple measurable function* if it takes finitely many values and $s^{-1}(\\set{e}) \\in \\Sigma$ for each $e \\in E$.\nThis extends the notion of a [[simple function|7WMBX7M4P6]] to values in $E$.\nA function $f \\colon X \\to E$ is *strongly measurable* if there is a sequence $(s_n)$ of simple measurable functions $s_n \\colon X \\to E$ with\n$$\\lim_{n \\to \\infty} \\norm{s_n(x) - f(x)} = 0 \\quad \\text{for every } x \\in X,$$\nthat is, $s_n \\to f$ [[pointwise|056FSKG32J]] on $X$.\n\n## Remarks\n\nExample: let $X = [0, 1]$ with the Borel σ-algebra and let $f \\colon X \\to E$ be continuous.\nThe function $s_n$ that equals $f(k/n)$ on $[k/n, (k+1)/n)$ for $0 \\le k \u003c n$ and $f(1)$ at $1$ is simple and measurable, and $s_n(x) \\to f(x)$ by continuity, so $f$ is strongly measurable.\n\nNon-example: let $X = [0, 1]$ with the σ-algebra $\\set{\\varnothing, X}$ and $f(x) = x$ with values in $E = \\R$.\nA simple measurable function is then constant, so a pointwise limit of such functions is constant, and $f$ is not strongly measurable.\n\nWhen $E = \\R$ or $E = \\C$, a function is strongly measurable if and only if it is measurable.\nFor 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.\nThe 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 $s_n$.\n",
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