Let \(P\) be a partition of \([a,b]\); for convenience, write it as

for \(a < x_1 < \cdots < x_{N-1} < b\). For even more convenience, let \(x_0 := a\) and \(x_N := b\). Now for any \(i \in \{1,\ldots,N\}\), the Mean Value Theorem implies that there is some \(\xi_i \in (x_{i-1},x_i)\) such that

Now trivially

so multiplying by the length of the interval \([x_{i-1},x_i]\) and summing over \(i = 1,\ldots,N\) gives that

The inner sum telescopes and equals \(f(x_N) - f(x_0) = f(b) - f(a)\). Therefore every lower sum is bounded above by \(f(b) - f(a)\) and every upper sum is bounded below by \(f(b) - f(a)\). This implies

and

Since we have assumed that \(f'\) is Riemann integrable, upper integral and lower integral are themselves equal and therefore must equal \(f(b) - f(a)\).

The second half of the Fundamental Theorem of Calculus reinforces strength of the assertion that differentiation and integration are inverse operations by switching the order relative to part 1. Although perhaps less frequently-used, this second part is also very important.

We already know that continuous functions are integrable, so the function \(g(x)\) given by

is well-defined for all \(x \in (a,b)\). Furthermore, we know that when \(y < y'\) are points in \((a,b)\),

Fixing \(y' := x+h\) and \(y := x\) for \(h > 0\) and alternately fixing \(y' := x\) and \(y := x+h\) when \(h < 0\) gives

when \(h > 0\) and

when \(h < 0\).

(Note that we have not technically defined the Riemann integral in the case when the bounds appear in the wrong order, which is why we don't write \(\int_{x}^{x+h}\) in the case \(h < 0\).) For any \(\epsilon > 0\), continuity of \(f\) at \(x\) implies the existence of some \(\delta > 0\) such that \(|f(t) - f(x) | < \epsilon\) for all \(t \in (a,b)\) with \(|t-x| < \delta\). If we suppose that \(|h| < \delta\), it follows that

for all \(t\) between \(x\) and \(x+h\), from which it follows that

when \(h > 0\) and

when \(h < 0\). Combining these observations gives

when \(0 < |h| < \delta\); therefore

Modulo the non-strict inequality (which is easily fixed by simply replacing \(\epsilon\) by \(\epsilon/2\), for example), this exactly implies that \(g\) is differentiable at \(x\) and has derivative \(f(x)\) there.

The proof is by an approximation argument. Let \(\varphi\) and \(f\) be as described. Suppose also for the moment that \(\varphi' > 0\). Given a partition \(P\) of \(\varphi([a,b])\), let \(f^+\) be a function which equals \(\sup_{x \in I} f(x)\) at each \(x \in I\) for each \(I \in P\) and let \(f^{-}\) be a function which equals \(\inf_{x \in I} f(x)\) at each \(x \in I\) for each \(I \in P\). There is technically some ambiguity in how \(f^+\) and \(f^-\) are defined at the endpoints of intervals (since such points belong to more than one \(I\)). For simplicity, define \(f^+(x)\) to be the maximum of \(\sup_{y \in I} f(y)\) for each \(I\) containing and likewise take \(f^{-}(x)\) to be the minimum of for each containing . Then \(f^-(x) \leq f(x) \leq f^+(x)\) for all \(x \in [a,b]\) and, aside from modification at finitely many points, \(f^{-}\) and \(f^+\) are linear combinations of characteristic functions of intervals. Thus

and

By linearity of the integral, it must be the case that

and likewise

Now because upper and lower integrals are monotone, we have

and

Because \(f\) is Riemann integrable, for any \(\epsilon\), there is some partition \(P\) making \(U(f,P) - L(f,P) < \epsilon\), so it must be the case that the upper and lower integrals of \(f(\varphi(t))|\varphi'(t)|\) on \([a,b]\) are in fact equal. Moreover, they must equal the integral of \(f\) on \(\varphi([a,b])\). If \(\varphi'\) is never negative, we are done. If not, \(|\varphi'(t)| = - \varphi'(t)\) and we use the definition of integration with reversed endpoints to finish the proof.