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  "body": "\n## Proof\n\nWe prove integrability of $f$ with Riemann's criterion, and then estimate the difference of the integrals.\n\n*Step A: integrable functions and $f$ are bounded.*\n\n1. Let $a \u003c b$, let each $f_n \\colon [a, b] \\to \\R$ be [[Riemann integrable|0N1JQXZXHC]], and let $f_n \\to f$ [[uniformly|08FPABDQ4C]] on $[a, b]$.\n2. First let $g \\colon [a, b] \\to \\R$ be any Riemann integrable function with integral $I$. Take $\\varepsilon = 1$ in the definition, and let $\\delta \u003e 0$ be the corresponding number.\n3. Fix a [[partition|069P20KYQT]] $P = (x_0, \\dots, x_m)$ of $[a, b]$ into equal pieces, with [[mesh|04GM4NYXJB]] less than $\\delta$.\n4. Suppose that $g$ is unbounded on $[a, b]$. Then $g$ is unbounded on some piece $[x_{j-1}, x_j]$, since a finite union of bounded sets is bounded.\n5. Fix tags $t_i \\in [x_{i-1}, x_i]$ for $i \\ne j$, and let $A = \\sum_{i \\ne j} g(t_i)(x_i - x_{i-1})$. For each $t \\in [x_{j-1}, x_j]$, the [[Riemann sum|0HBY48G8G1]] with tag $t$ on the $j$-th piece is $A + g(t)(x_j - x_{j-1})$.\n6. By step 2, this sum lies within $1$ of $I$. So $|g(t)| \\le \\frac{|I| + 1 + |A|}{x_j - x_{j-1}}$ for all $t \\in [x_{j-1}, x_j]$. This contradicts step 4, so $g$ is bounded.\n7. By uniform convergence, there is $N_1$ with $|f_{N_1}(x) - f(x)| \u003c 1$ for all $x \\in [a, b]$. By step 6, there is $K$ with $|f_{N_1}| \\le K$. So $|f| \\le K + 1$, and $f$ is bounded.\n\n*Step B: $f$ is integrable.*\n\n8. Let $\\varepsilon \u003e 0$, and put $\\eta = \\frac{\\varepsilon}{4(b - a)}$. By uniform convergence, there is $N$ with $|f_n(x) - f(x)| \u003c \\eta$ for all $n \\ge N$ and all $x \\in [a, b]$.\n9. The function $f_N$ is integrable and bounded. By [[Riemann's criterion|017DH8WXTE]], there is a partition $P = (x_0, \\dots, x_m)$ with $U(f_N, P) - L(f_N, P) \u003c \\frac{\\varepsilon}{2}$. Here $U$ and $L$ are the [[upper and lower Darboux sums|0QRQVH0CRW]].\n10. Fix $i$, and let $J = [x_{i-1}, x_i]$. For $x \\in J$, step 8 gives $f(x) \u003c f_N(x) + \\eta \\le \\sup f_N(J) + \\eta$. So $\\sup f(J) \\le \\sup f_N(J) + \\eta$, by the definition of the [[supremum|0BTMQ04556]].\n11. In the same way, $f(x) \u003e f_N(x) - \\eta \\ge \\inf f_N(J) - \\eta$ for $x \\in J$. So $\\inf f(J) \\ge \\inf f_N(J) - \\eta$, by the definition of the [[infimum|04KFTA34C5]].\n12. Multiply steps 10 and 11 by $x_i - x_{i-1}$ and sum over $i$. Since the lengths add up to $b - a$,\n$$U(f, P) - L(f, P) \\le U(f_N, P) - L(f_N, P) + 2 \\eta (b - a) \u003c \\frac{\\varepsilon}{2} + \\frac{\\varepsilon}{2} = \\varepsilon.$$\n13. Since $\\varepsilon \u003e 0$ was arbitrary and $f$ is bounded by step 7, Riemann's criterion shows that $f$ is Riemann integrable on $[a, b]$.\n\n*Step C: the integrals converge.*\n\n14. Let $\\varepsilon \u003e 0$, put $\\eta = \\frac{\\varepsilon}{2(b - a)}$, and let $N$ be as in step 8 for this $\\eta$. Fix $n \\ge N$.\n15. By step 13 and [[linearity of the integral|080RWVVSB6]], $f_n - f$ is integrable, and $\\int_a^b f_n - \\int_a^b f = \\int_a^b (f_n - f)$.\n16. By the [[triangle inequality for integrals|0NR82KMYKD]], $|f_n - f|$ is integrable and $\\big| \\int_a^b (f_n - f) \\big| \\le \\int_a^b |f_n - f|$.\n17. The constant function $\\eta$ has integral $\\eta (b - a)$, by the [[integral of a power|0Q3HYFT9RQ]] with exponent $0$ and linearity. Since $|f_n - f| \\le \\eta$, [[monotonicity of the integral|0MBWQKM8QR]] gives $\\int_a^b |f_n - f| \\le \\eta (b - a) = \\frac{\\varepsilon}{2}$.\n18. By steps 15 to 17, $\\big| \\int_a^b f_n - \\int_a^b f \\big| \\le \\frac{\\varepsilon}{2} \u003c \\varepsilon$ for all $n \\ge N$.\n19. So $\\lim_{n \\to \\infty} \\int_a^b f_n(x) \\, dx = \\int_a^b f(x) \\, dx$, where the right side is the [[Riemann integral|0QM4TXT3DR]] of $f$.\n\n## Idea\n\nUniform convergence traps the graph of $f$ in a thin band of width $2\\eta$ around the graph of $f_N$. The upper and lower sums of $f$ can then differ by at most $2\\eta(b - a)$ more than those of $f_N$. The same band bounds the difference of the integrals by $\\eta(b - a)$.\n",
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