update notations and fix typos
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@@ -40,7 +40,7 @@ $$
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Therefore, $U(P,f,\alpha) - L(P,f,\alpha)<U(P,\alpha,f) - L(P,\alpha,f) < \epsilon$, so $f\in \mathscr{R}(\alpha)$ on $[a, b]$.
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EOP
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QED
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#### Theorem 6.10
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@@ -76,7 +76,7 @@ Since $\epsilon$ is arbitrary, we have $U(P,f,\alpha) - L(P,f,\alpha) < \epsilon
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Therefore, $f\in \mathscr{R}(\alpha)$ on $[a, b]$.
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EOP
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QED
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#### Theorem 6.11
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