1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Is $g$ assumed to be differentiable everywhere? $\endgroup$ – Bungo Jul 11 '19 at 17:12
  • $\begingroup$ Differentiable on $[c,d]$ $\endgroup$ – Dendrilo Jul 11 '19 at 17:43
5
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.