Fixed typo in FTC for Riemann integrals.
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@@ -95,10 +95,10 @@
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\end{enumerate}
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\end{enumerate}
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\end{theorem}
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\end{theorem}
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\begin{proof}
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\begin{proof}
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(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 $h > 0$,
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(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 $h > 0$,
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\begin{align*}
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\begin{align*}
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\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)))} \\
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\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)))} \\
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-\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)))}
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-\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]))}
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\end{align*}
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\end{align*}
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As $E$ is locally convex and separated, and $f$ is continuous at $x$, this implies that $[F(x + h) - F(x)]/h \to f(x)$.
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As $E$ is locally convex and separated, and $f$ is continuous at $x$, this implies that $[F(x + h) - F(x)]/h \to f(x)$.
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