Does monotonic function imply continuity?

Question

My formula book states that in order to apply variable substitution on an integral, i.e use the formula: $$ \int_a^b f(x)dx = \int_{g^{-1}(a)}^{g^{-1}(b)} f(g(t))g'(t)dt $$ $g(t)$ needs to be a monotonic function in $g^{-1}(a) \leq t \leq g^{-1}(b)$. However, the substitution works for example for the Gamma function: $$ \Gamma(z) = \int_0^ \infty u^{z-1}e^{-u}du $$ Applying the substitution $u = g(t) = \ln{\frac{1}{t}}$ in $\Gamma (z)$, one obtains $$ \Gamma(z) = \int_0^1 [\ln (\frac{1}{t})]^{z-1}dt $$ which also seem to give correct results (ref: mathworld.wolfram.com). But there is a discontinuity in $0$. My formula book begins stating the conditions for a monotonic function for a continuous function in the closed interval $[x,y]$. I interpret this as one condition for monotonicity is that the function first of all be continuous in the closed interval. Or can a function be monotonic without being continuous?

Answer 1

I think there's two related points to make to answer your question:

1. A function can certainly be monotonic but not continuous. Consider $$H(x) = \begin{cases} 0 & x \leq 0 \\ 1 & \text{otherwise} \end{cases}$$

2. If one thinks of substituions as $u = f(x)$, then $f$ should usually at least be continuous and monotonic on the open interval of your integration $(a, b)$. If there is a discontinuity only at an endpoint, then often things "just work out" if you treat it as normal. If there is a discontinuity at the endpoint, then strictly, one should consider an interval $[a + \epsilon, b]$ and take the limit $\epsilon \to 0$.

Generally, the monotonicity is required so that you don't "lose" part of your interval. For example, consider what would happen if we blindly went ahead and tried to substitute $u = x^2$ in a rather trivial integral $$I = \int_{-1}^1 dx = \int_1^1 \frac{1}{2\sqrt{u}} \; du$$ We've ended up with $1$ at both endpoints of our interval - this is complete rubbish! In much the same way, you'd never try to substitute $u = 1$ either - it's not a good parametrisation (in both the formal and informal senses) of your region of integration.

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Edit: Raoul correctly points out in the comment thread below that monotonicity is not necessary whatsoever to perform substitutions of the form $x = f(u)$. One should consider my answer more in the context of $u = f(x)$, where we exploit monotonicity to provide an inverse (recasting the problem as x = f^{-1}(u)$, and then defer to Raoul's comment (a formally correct statement of "integration by substitution").