# Integration by substitution's theorem only with $f$ Riemann integrable and $g$ a monotone function.

I'm trying to solve this problem: Let $$f\colon\left[a,b\right]\to\mathbb{R}$$ be a Riemann integrable function and $$g\colon\left[c,d\right]\to\mathbb{R}$$ be a monotone function such that $$g'$$ is Riemann integrable. Prove that if $$g\left(\left[c,d\right]\right)\subseteq\left[a,b\right]$$, then $$\int_{g\left(c\right)}^{g\left(d\right)}f\left(x\right)dx=\int_{c}^{d}f\left(g\left(t\right)\right)\cdot g'\left(t\right)dt$$.

This is similar to the integration by substitution's theorem, but in that theorem is given the hypothesis that $$f$$ is continuous and therefore it has antiderivative $$F\left(x\right)=\int_{a}^{x}f\left(t\right)dt$$ and the prove is easy applying the change rule to the function $$F\left(g\left(t\right)\right)$$. But I don't know how to use the fact that $$g$$ is monotone in the problem, because in this case we only know that $$f$$ is Riemann integrable and may be it has no antiderivative. Could you help me or give me some suggestions?

Thanks.

• Is $g$ assumed to be differentiable everywhere? – Bungo Jul 11 '19 at 17:12
• Differentiable on $[c,d]$ – Dendrilo Jul 11 '19 at 17:43

Assuming only that $$f$$ is Riemann integrable and $$g$$ is differentiable and monotone, this is straightforward to prove with the additional condition that $$g'$$ is continuous on $$[c,d]$$.

We assume without loss of generality that $$g$$ is non-decreasing.

Consider a partition $$P: c = x_0 < x_1 < \ldots < x_n = d$$ and the Riemann sum

$$\tag{1}S(P,fg')= \sum_{j=1}^n f(g(\xi_j))g'(\xi_j)(x_j - x_{j-1})$$

using intermediate points $$\xi_j \in [x_{j-1},x_j]$$.

If $$g$$ is increasing, then we have a partition $$P'$$ of $$[g(c),g(d)]$$ given by

$$g(c) = g(x_0) < g(x_1) < \ldots < g(x_n) = g(d),$$

Using the intermediate points $$g(\xi_j)$$, we have a Riemann sum for the integral of $$f$$ over $$[g(c),g(d)]$$ taking the form

$$S(P',f) = \sum_{j=1}^n f(g(\xi_j))(\,g(x_j) - g(x_{j-1})\,),$$

where the monotonicity of $$g$$ is needed to ensure that $$g(\xi_j) \in [g(x_{j-1}), g(x_j)]$$.

Applying the mean value theorem, there exist points $$\eta_j \in (x_{j-1},x_j)$$ such that

$$\tag{2}S(P',f) = \sum_{j=1}^n f(g(\xi_j))g'(\eta_j)(x_j - x_{j-1})$$

Thus,

$$\tag{3}\left|\int_{g(c)}^{g(d)} f(x) \, dx - S(P,fg') \right| \leqslant \left|\int_{g(c)}^{g(d)} f(x) \, dx - S(P',f) \right|+ \left|S(P',f) - S(P,fg') \right|$$

Since $$f$$ is Riemann integrable, for any $$\epsilon > 0$$ there exists $$\delta_1$$ such that if $$\|P'\| < \delta_1$$, then the first term on the RHS of (3) is less than $$\epsilon/2$$.

Using (1) and (2), we also have for the second term on the RHS of (3),

$$\left|S(P',f) - S(P,fg') \right| = \left|\sum_{j=1}^n f(g(\xi_j))\left(\, g'(\xi_j)-g'(\eta_j)\,\right)(x_j - x_{j-1}) \right| \\ \leqslant \sum_{j=1}^n| f(g(\xi_j))|\left|\, g'(\xi_j)-g'(\eta_j)\,\right|(x_j - x_{j-1})$$

Since $$f\circ g$$ is bounded and $$g'$$ is continuous and, hence, uniformly continuous on $$[c,d]$$, we can find $$\delta_2 > 0$$ such that if $$\|P\| < \delta_2$$ then $$\left|S(P',f) - S(P,fg') \right| < \epsilon/2$$. Also by continuity of $$g$$ there exists $$\delta_3$$ such that if $$\|P\| < \delta_3$$, then $$\|P'\| < \delta_1$$.

Therefore, if $$\|P\| < \min(\delta_2, \delta_3)$$ then it follows that

$$\left|\int_{g(c)}^{g(d)} f(x) \, dx - S(P,fg') \right|< \epsilon,$$

which proves that $$(f\circ g) g'$$ is integrable and

$$\int_{g(c)}^{g(d)} f(x) \, dx= \int_{c}^{d} f(g(t))g'(t) \, dt$$

This can also be proved more generally if $$g'$$ is not assumed to be continuous, but it is more difficult. A reference is H. Kestelman, Mathematical Gazette, 45 [1961], 17-33.