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  "body": "\n## Proof\n\nWe first show that $|f|$ is integrable by Riemann's criterion, and then compare integrals.\n\n1. Let $a \u003c b$ and let $f \\colon [a, b] \\to \\mathbb{R}$ be [[Riemann integrable|0N1JQXZXHC]] with integral $I$. We first show that $f$ is bounded.\n   Take $\\delta \u003e 0$ such that each tagged partition of mesh less than $\\delta$ has [[Riemann sum|0HBY48G8G1]] within $1$ of $I$.\n2. By the [[Archimedean property|0A3GWZ86AY]], fix a partition $P = (x_0, \\dots, x_n)$ with all subintervals shorter than $\\delta$, and fix tags $t$.\n3. Fix $i$ and $x \\in [x_{i-1}, x_i]$. Let $t'$ be the tags $t$ with $t_i$ replaced by $x$. Both sums are within $1$ of $I$, so their difference $(f(x) - f(t_i))(x_i - x_{i-1})$ has absolute value less than $2$.\n   So $|f(x)| \u003c |f(t_i)| + \\frac{2}{x_i - x_{i-1}}$. The largest of these $n$ bounds bounds $f$ on $[a, b]$.\n4. Now let $Q$ be any partition and $J$ one of its subintervals. Let $M$ and $m$ be the supremum and infimum of $f$ on $J$, and $M'$ and $m'$ those of $|f|$.\n5. For $x, y \\in J$, both $f(x)$ and $f(y)$ lie in $[m, M]$. By the [[reverse triangle inequality|0C9S13WKMF]], $|f(x)| - |f(y)| \\le |f(x) - f(y)| \\le M - m$.\n6. Fix $y \\in J$. Step 5 says that $|f(y)| + (M - m)$ is an upper bound of $|f|$ on $J$, so $M' \\le |f(y)| + (M - m)$.\n   So $M' - (M - m)$ is a lower bound of $|f|$ on $J$, and $M' - (M - m) \\le m'$. Hence $M' - m' \\le M - m$.\n7. Multiply by the length of $J$ and add over the subintervals of $Q$. For the [[Darboux sums|0QRQVH0CRW]], $U(|f|, Q) - L(|f|, Q) \\le U(f, Q) - L(f, Q)$.\n8. Let $\\varepsilon \u003e 0$. By [[Riemann's criterion|017DH8WXTE]] for the bounded integrable $f$, there is $Q$ with $U(f, Q) - L(f, Q) \u003c \\varepsilon$. By step 7, the same $Q$ works for $|f|$.\n   Since $|f|$ is bounded, Riemann's criterion shows that $|f|$ is Riemann integrable.\n9. By the [[linearity of the integral|080RWVVSB6]] with coefficients $-1$ and $0$, $-|f|$ is integrable with integral $-\\int_a^b |f(x)| \\, dx$.\n10. By the definition of the [[absolute value|00S7C68D51]], $-|f(x)| \\le f(x) \\le |f(x)|$ for all $x$. By the [[monotonicity of the integral|0MBWQKM8QR]],\n    $$-\\int_a^b |f(x)| \\, dx \\le \\int_a^b f(x) \\, dx \\le \\int_a^b |f(x)| \\, dx.$$\n11. This says $\\big| \\int_a^b f(x) \\, dx \\big| \\le \\int_a^b |f(x)| \\, dx$.\n\n## Idea\n\nTaking absolute values can only shrink the spread of values on a subinterval. So the gap between upper and lower sums of $|f|$ is at most that of $f$, and $|f|$ is integrable. The inequality then follows by integrating $-|f| \\le f \\le |f|$.\n",
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