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  "body": "\n## Proof\n\nWe compare $X$ with a multiple of an indicator function.\n\n1. Let $X$ be a random variable with values in $[0, \\infty)$ on a probability space $(\\Omega, \\mathcal{F}, P)$, and let $a \u003e 0$.\n2. By the definition of the [[expected value|0DCT0QZVQ1]], $E[X] \\in [0, \\infty]$. If $E[X] = \\infty$, the right side is $\\infty$ and the inequality holds. So assume $E[X] \u003c \\infty$.\n3. Let $A = \\set{X \\ge a}$, an event, and let $\\mathbf{1}_A$ be its [[indicator function|0VSB0XF4G7]].\n4. We claim $a \\mathbf{1}_A \\le X$ pointwise. On $A$, the left side is $a$ and $X \\ge a$. Off $A$, the left side is $0$ and $X \\ge 0$.\n5. The integral of an indicator is the probability of its set, so $E[a \\mathbf{1}_A] = a P(A)$.\n6. By step 4, $X - a \\mathbf{1}_A \\ge 0$. The integral of a nonnegative function is nonnegative, so $E[X - a \\mathbf{1}_A] \\ge 0$.\n7. Both $X$ and $a \\mathbf{1}_A$ are integrable. By [[linearity of expectation|0P7PTC8DDE]] and step 5, $E[X] - a P(A) \\ge 0$.\n8. Dividing by $a \u003e 0$ gives $P(X \\ge a) \\le E[X]/a$.\n\n## Idea\n\nThe variable $X$ is at least $a$ on the event $\\set{X \\ge a}$ and at least $0$ elsewhere. So $X$ dominates $a$ times the indicator of that event. Taking expectations turns this pointwise bound into the inequality.\n",
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