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  "body": "\n## Definition\n\nLet $(X, \\Sigma, \\mu)$ be a [[measure space|P15VQS5DJ9]].\nLet $E$ be a [[Banach space|0BDTTM24DP]] over $\\R$ or $\\C$ with [[dual space|0M856Q1AMZ]] $E^*$.\nLet $f \\colon X \\to E$ be a function such that $\\varphi \\circ f$ is [[integrable|X7GTF3D98F]] for every $\\varphi \\in E^*$.\nThe function $f$ is *Pettis integrable* if for every $A \\in \\Sigma$ there is a vector $x_A \\in E$ with\n$$\\varphi(x_A) = \\int_A \\varphi \\circ f \\, d\\mu \\quad \\text{for every } \\varphi \\in E^*,$$\nwhere the right side is a [[Lebesgue integral|DWJF7ABPR1]].\nThe vector $x_A$ is unique, and it is the *Pettis integral* $\\int_A f \\, d\\mu$.\n\n## Remarks\n\nThe vector $x_A$ is unique because the continuous linear functionals separate the points of $E$, by the Hahn–Banach theorem.\n\nExample: let $E = \\R^n$ and let each component $f_k$ of $f$ be integrable.\nThen $x_A = (\\int_A f_1 \\, d\\mu, \\dots, \\int_A f_n \\, d\\mu)$ works, because every $\\varphi \\in E^*$ is a linear combination of the coordinates.\n\nNon-example: let $X = \\N$ with the counting measure, $E = c_0$, the space of sequences tending to $0$, and $f(n) = e_n$, the $n$-th unit vector.\nFor each $\\varphi \\in E^* = \\ell^1$, the sum $\\sum_n \\abs{\\varphi(e_n)}$ is finite, so $\\varphi \\circ f$ is integrable.\nBut for $A = \\N$ the only candidate is the constant sequence $(1, 1, \\dots)$, which is not in $c_0$, so $f$ is not Pettis integrable.\n\nEvery Bochner integrable function is Pettis integrable, with the same integral.\nIn infinite-dimensional spaces the converse fails.\n",
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