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  "body": "\n## Statement\n\nLet $(X, \\Sigma)$ be a measurable space and let $E$ be a Banach space over $\\R$ or $\\C$ with [[dual space|0M856Q1AMZ]] $E^*$.\nLet $f \\colon X \\to E$ be a function.\nThen $f$ is [[strongly measurable|JDM2M1FRY6]] if and only if both of the following hold.\n\n1. The function $\\varphi \\circ f$ is [[measurable|YYNCJSR0PJ]] for every continuous linear functional $\\varphi \\in E^*$, that is, $f$ is *weakly measurable*.\n2. The image $f(X)$ is a [[separable|0GKJ78948V]] subset of $E$.\n\n## Remarks\n\nThe direct implication is easy: $\\varphi \\circ s_n \\to \\varphi \\circ f$ pointwise, and $f(X)$ lies in the closure of the countably many values of the approximating simple functions $s_n$.\nFor the converse, one may replace $E$ by the closed linear span of $f(X)$, which is separable.\nIn a separable Banach space there is a sequence $(\\varphi_k)$ in $E^*$ with $\\norm{e} = \\sup_k \\abs{\\varphi_k(e)}$ for all $e$.\nSo $x \\mapsto \\norm{f(x) - e}$ is measurable for each $e$, and one builds simple approximations by sending $x$ to the nearest point of a countable dense set among its first $n$ points.\n\nWhen $E$ is separable, condition 2 is automatic, so weak and strong measurability coincide.\nWithout condition 2 the equivalence fails.\nTake $X = [0, 1]$ with the Borel σ-algebra and the map $f(t) = \\mathbf{1}_{\\set{t}}$ into the non-separable Hilbert space $\\ell^2([0, 1])$.\nIt is weakly measurable, since $\\varphi \\circ f$ vanishes outside a countable set for each $\\varphi$.\nBut $f(X)$ is not separable, so $f$ is not strongly measurable.\n\nThere is also a version relative to a measure $\\mu$.\nThen $f$ is strongly $\\mu$-measurable if and only if it is weakly $\\mu$-measurable and takes values in a separable subset of $E$ outside a $\\mu$-null set.\n\n## Explanations\n\n### L1 human\n\nTesting measurability of a vector-valued function directly is hard.\nThe theorem says that it is enough to test each linear functional, one at a time.\nThis works as long as the values of the function stay inside a separable part of the space.\nSo questions about a function in a large space reduce to questions about ordinary real or complex functions.\n\n### L2 human\n\nThe theorem is the main tool to check the hypotheses of the Bochner integral.\nWith Bochner's criterion, it shows when $f$ is Bochner integrable.\nThis holds as soon as $f$ is weakly measurable, essentially separably valued and $\\int \\norm{f} \\, d\\mu \u003c \\infty$.\nFor a separable space $E$, weak measurability alone suffices.\nThe theorem also explains why the Pettis integral, which only asks for weak measurability, covers more functions than the Bochner integral in non-separable spaces.\n",
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