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  "body": "\n## Definition\n\nLet $U$ be a nonempty [[compact set|0R6EWFKRCF]] in $\\R^m$, and let $\\mathcal{P}(U)$ be the set of Borel [[probability measures|059TJN0NZ1]] on $U$.\nLet $T \u003e 0$.\nA *relaxed control* on $[0, T]$ is a family $(\\mu_t)_{t \\in [0, T]}$ with $\\mu_t \\in \\mathcal{P}(U)$ for every $t$.\nIt must be measurable in this sense: for every continuous function $h \\colon U \\to \\R$, the map $t \\mapsto \\int_U h \\, d\\mu_t$ is Lebesgue measurable on $[0, T]$.\nFor a [[control system|ZPDTVZQN5K]] $x' = f(t, x, u)$ with $f$ continuous, the trajectory of the relaxed control $(\\mu_t)$ solves the relaxed equation\n$$x'(t) = \\int_U f(t, x(t), v) \\, d\\mu_t(v).$$\n\n## Remarks\n\nEvery ordinary [[admissible control|8M31QKB0GW]] $u \\colon [0, T] \\to U$ gives the relaxed control $\\mu_t = \\delta_{u(t)}$, the Dirac measure at $u(t)$.\nIts relaxed equation is $x' = f(t, x, u(t))$, the original one.\nExample: $U = \\set{-1, 1}$ and $\\mu_t = \\frac12 \\delta_{-1} + \\frac12 \\delta_1$ for all $t$.\nIt is a relaxed control, since $\\int_U h \\, d\\mu_t = \\frac12 (h(-1) + h(1))$ is constant in $t$.\nFor $f(t, x, u) = u$ its trajectory has $x' = 0$, a velocity that no ordinary control with values in $\\set{-1, 1}$ gives.\nNon-example: $\\mu_t = \\delta_{-1} + \\delta_1$ is not a relaxed control, since $\\mu_t(U) = 2$, so $\\mu_t$ is not a probability measure.\nThe velocities of relaxed controls fill the convex hull of $f(t, x, U)$, so the convexity hypothesis of [[Filippov's existence theorem|XEX7QZV1BS]] holds for the relaxed problem.\nA relaxed control is a [[Young measure|6TQSVJQDEH]] on $[0, T]$ with values in $U$.\n",
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