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  "body": "\n## Statement\n\nLet $F$ be a [[cumulative distribution function|0H8H7ZTHW2]] with $F(0) = 0$, so that $F$ is the law of a positive random variable.\nAssume that $F$ is not arithmetic, that is, $F$ is not concentrated on a set $\\set{0, h, 2h, \\dots}$ for any $h \u003e 0$.\nLet $\\mu = \\int_{[0, \\infty)} t \\, dF(t)$ be its mean, with $0 \u003c \\mu \\le \\infty$.\nLet $g \\colon [0, \\infty) \\to \\R$ be directly Riemann integrable.\nThis means that, for the partition of $[0, \\infty)$ into intervals of length $h$, the upper and lower Riemann sums of $g$ are finite.\nIt also means that they tend to the same limit as $h \\to 0$.\nConsider the renewal equation\n$$u(t) = g(t) + \\int_{[0, t]} u(t - s) \\, dF(s), \\qquad t \\ge 0,$$\nwhere the integral is a [[Lebesgue–Stieltjes integral|FK3QBZAZH7]].\nThen:\n\n1. The equation has exactly one solution $u$ that is bounded on bounded intervals, namely $u(t) = \\int_{[0, t]} g(t - s) \\, dU(s)$. Here $U = \\sum_{n \\ge 0} F^{*n}$, where $F^{*n}$ is the $n$-fold convolution power of $F$ and $F^{*0}$ is the unit mass at $0$.\n2. As $t \\to \\infty$,\n$$u(t) \\to \\frac{1}{\\mu} \\int_0^\\infty g(s) \\, ds,$$\nwhere $1/\\mu$ is read as $0$ when $\\mu = \\infty$.\n\n## Remarks\n\nPart 1 follows by iterating the equation: the series of convolution powers converges because $F^{*n}(t) \\to 0$ fast for each fixed $t$.\nPart 2 is the key renewal theorem.\nIts proof first treats $g$ equal to the indicator function of $[0, h)$, which is Blackwell's renewal theorem.\nIt then approximates a directly Riemann integrable $g$ from above and below by step functions.\nA nonnegative, nonincreasing, Lebesgue integrable function on $[0, \\infty)$ is directly Riemann integrable.\n\nExample: for $g = 1 - F$, the constant $u = 1$ solves the equation, since $1 - F(t) + F(t) = 1$.\nThe limit in part 2 is $\\frac{1}{\\mu} \\int_0^\\infty (1 - F(s)) \\, ds = \\mu / \\mu = 1$, as it must be.\n\nThe hypothesis that $F$ is not arithmetic is needed.\nIf $F$ is the unit mass at $1$, then $U$ puts mass $1$ on each integer $n \\ge 0$.\nFor $g$ the indicator function of $[0, 1/2)$, the solution is $u(t) = 1$ when the fractional part of $t$ is less than $1/2$ and $u(t) = 0$ otherwise, so $u(t)$ has no limit.\n\n## Explanations\n\n### L1 human\n\nA renewal equation describes a quantity that is fed by a source $g$ and by its own past, where the past is weighted by a waiting time law $F$.\nThe theorem says that after a long time the quantity settles down to a constant.\nThe constant is the total size of the source divided by the mean waiting time $\\mu$.\nFor example, the expected number of light bulb replacements in a time window of length $h$ tends to $h / \\mu$.\n\n### L2 human\n\nIn [[renewal theory|45V0W838BQ]], $U(t)$ is the expected number of renewals in $[0, t]$, counting the start.\nMany quantities of a renewal process, such as the law of the residual waiting time at time $t$, satisfy a renewal equation with an explicit $g$.\nThe theorem then gives their limits as $t \\to \\infty$.\nThe same method describes the long-time behaviour of age-structured populations through the [[renewal equation|EWVVH3EKMR]] and the [[Euler–Lotka equation|E0YVMBPK7A]], after a change of variable that makes $F$ a probability distribution.\nFrom the point of view of [[Volterra integral equations|DND8CBEVXT]], it is a statement about the [[Volterra equation of convolution type|SNFYTC1TTA]] whose kernel is a probability measure.\n",
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