From 95829261c79b1cca25ac94aefc8e2111b1d510c6 Mon Sep 17 00:00:00 2001 From: Bokuan Li Date: Tue, 28 Apr 2026 18:09:27 -0400 Subject: [PATCH] Fixed typo in FTC for Riemann integrals. --- src/fa/rs/regulated.tex | 6 +++--- 1 file changed, 3 insertions(+), 3 deletions(-) diff --git a/src/fa/rs/regulated.tex b/src/fa/rs/regulated.tex index 624d8c1..b851aa6 100644 --- a/src/fa/rs/regulated.tex +++ b/src/fa/rs/regulated.tex @@ -95,10 +95,10 @@ \end{enumerate} \end{theorem} \begin{proof} - (1): Let $x_0 \in (a, b)$ such that $f$ is continuous at $x$, and $\delta > 0$ such that $(x_0 - \delta, x_0 + \delta) \subset (a, b)$, then for any $h > 0$, + (1): Let $x \in (a, b)$ such that $f$ is continuous at $x$, and $\delta > 0$ such that $(x_0 - \delta, x_0 + \delta) \subset (a, b)$, then for any $h > 0$, \begin{align*} - \frac{1}{h}\braks{\int_a^{x+h} f(t)dt - \int_a^{x} f(t)dt} &= \frac{1}{h}\int_x^{x+h}f(t)dt \in \overline{\text{Conv}(f([t, t +h)))} \\ - -\frac{1}{h}\braks{\int_a^{x-h} f(t)dt - \int_a^{x} f(t)dt} &= \frac{1}{h}\int_{x-h}^{x}f(t)dt \in \overline{\text{Conv}(f([t, t +h)))} + \frac{1}{h}\braks{\int_a^{x+h} f(t)dt - \int_a^{x} f(t)dt} &= \frac{1}{h}\int_x^{x+h}f(t)dt \in \overline{\text{Conv}(f([x, x +h)))} \\ + -\frac{1}{h}\braks{\int_a^{x-h} f(t)dt - \int_a^{x} f(t)dt} &= \frac{1}{h}\int_{x-h}^{x}f(t)dt \in \overline{\text{Conv}(f((x - h, x]))} \end{align*} As $E$ is locally convex and separated, and $f$ is continuous at $x$, this implies that $[F(x + h) - F(x)]/h \to f(x)$.