# Question regarding integration by substitution and Riemann sums

I was reading Advanced Calculus: A Geometric View and I liked how he explained in integration by substitution why the differential changes the way it does(both in push-forward and pullback substitutions(that is, direct and inverse substitution)).

However there is a step that he didn't verify(at least as per my understanding) which I couldn't prove, that is:

In pullback substitution, consider the function $$y = f(x)$$ and $$x = g(s)$$, then the statement of pullback substitution for definite integral is: $$\int_{a}^{b}f(x)dx = \int_{g^{-1}(a)}^{g^{-1}(b)}f(g(s))g'(s)ds$$ He justified this using Riemann sums: In the simplest case a left-endpoint Riemann sum with $$n$$ equal subintervals , set $$\Delta{x} = (b-a)/n$$, set $$x_{i} = a + (i-1)\Delta{x}$$, then

$$\int_{a}^{b}f(x)dx = \lim_{n \to \infty} \sum_{i=n}^{n} f(x_{i}) \Delta{x}$$

And since $$\lim_{s \to s_{0}} \frac{g(s) - g(s_{0})}{s-s_{0}} = \lim_{\Delta{s} \to 0} \frac{\Delta{x}}{\Delta{s}} = g'(s_{0}) \implies \Delta{x} \approx g'(s_{0})\Delta{s} \text{, when } \Delta{s} \approx 0 \text{ (that is, the linear approximation of g at s_{0})}$$

$$\text{So we have:}\sum_{i=n}^{n} f(x_{i}) \Delta{x} \approx \sum_{i=n}^{n} f(g(s_{i})) g'(s_{i}) \Delta{s_{i}}$$

Where generally the subintervals $$\Delta{s_{i}}$$ are not equal, and in the limit when $$n \to \infty$$ both Riemann sums becomes an integral.

What the book said then is the following:

"By choosing $$n$$ sufficiently large, we can make every $$\Delta{s_{i}}$$ arbitrarily small and thus can make these two sums arbitrarily close"

I don't fully understand this statement, and how it can be used to prove that indeed in the limit when $$n \to \infty$$ both Riemann sums approaches the same limit(that is, the same integral).

Edit

Also how could we justify all this in the general Riemann sum, where the subintervals are not of equal length and the sampling point chosen randomly in each subinterval? As in this case we will have a summation that involves two different sampling point in each subinterval: $$\sum f(g(\bar{s_{i}})) g'(s_{i}) \Delta{s_{i}}$$

• Yeah this isn't especially rigorous. The theorem holds if $g$ is a $C^1$ diffeomorphism Commented Dec 3, 2023 at 11:26
• @FShrike where can I read more about this ? note that in the book it mentioned that $g$ must be invertible as well. Commented Dec 3, 2023 at 11:49
• diffeomorphism means, in particular, invertible (as well as some other things) Commented Dec 3, 2023 at 12:15
• @FShrike but what then is the proof that the two limits are equal? Commented Dec 3, 2023 at 12:20

Very briefly (I don't have time to elaborate at the moment). Let $$g$$ be a $$C^1$$ diffeomorphism $$[a,b]\to[c,d]$$.

Let $$f:[c,d]\to\Bbb R$$ be a Riemann integrable function. Assume $$(f\circ g)\cdot g'$$ is also Riemann integrable.

Let $$(x_0,\cdots,x_n),(t_1,\cdots,t_n)$$ be a tagged partition of $$[c,d]$$. We consider the Riemann sum $$\sum_{j=1}^nf(t_j)(x_j-x_{j-1})$$. Because $$g$$ is invertible we know in particular the $$t_j$$ pullback to unique $$s_j$$ and the $$x_y$$ pullback to unique $$u_j$$ and we find: $$\sum_{j=1}^nf(t_j)(x_j-x_{j-1})=\sum_{j=1}^nf(g(s_j))(g(u_j)-g(u_{j-1}))=\sum_{j=1}^nf(g(s_j))g'(\xi_j)(u_j-u_{j-1})$$For some $$\xi_j\in(u_{j-1},u_j)$$, by the mean value theorem.

The $$(u_\bullet),(s_\bullet)$$ will also form a tagged partition of $$[a,b]$$ up to a reordering (why? What do we know about continuous invertible functions on the real line?). There will be two cases but we can ignore this ambiguity - why?

$$g'(\xi_j)=g'(s_j)+g'(\xi_j)-g'(s_j)$$ and that sum becomes: $$\sum_{j=1}^nf(g(s_j))g'(s_j)(u_j-u_{j-1})+\sum_{j=1}^nf(g(s_j))(g'(\xi_j)-g'(s_j))(u_j-u_{j-1})$$The first term is a Riemann sum for $$(f\circ g)g'$$. We can use the continuity of $$g^{-1}$$ (how?) to show that the mesh of the partition $$(u_\bullet)$$ is controllable in terms of the mesh of the partition $$(x_\bullet)$$ so that this first term can become arbitrarily close to $$\int_c^d f(g(u))g'(u)\,\mathrm{d}u$$. We then want to remove the error term in the limit (and to be precise, what limit are we actually taking? That depends somewhat on how you view the Riemann integral... it's actually better to live in $$\epsilon$$-$$\delta$$ land for this and avoid explicit limits in my opinion).

We use the fact Riemann integrable functions are (by the classical definition) bounded and the fact $$g'$$ is continuous; note $$\xi_j$$ can be made really very close to $$u_j$$. We actually need more than that though, don't we? We require uniform continuity. Luckily, that's for free: if $$h:[q,r]\to\Bbb R$$ is continuous it is also uniformly continuous. You should be able to show that the error term is bounded by $$M\epsilon(b-a)$$ where $$M$$ is a bound for $$f$$ and $$\epsilon$$ is some arbitrarily small constant, assuming we chose the mesh of $$x_\bullet$$ to be small enough!

• Thanks for this outline, where are these stuff typically explained? as I didn't encounter it in Calculus Commented Dec 3, 2023 at 12:40
• @LoaiGhoraba That's because this stuff is really real analysis, and many introductory calculus texts brush over it. I think a good real analysis text or a more rigorous calculus text - e.g. Spivak's calculus gets rave reviews - would be good for you. There are plenty freely available Commented Dec 3, 2023 at 12:41
• Thanks a lot! That was really helpful. Commented Dec 3, 2023 at 12:48
• @LoaiGhoraba Oh, but the MVT is linear approximation. It's just a more explicit form! All differentiation is is linear approximation. In saying that $g(u_j)-g(u_{j-1})=g'(\xi)(u_j-u_{j-1})$, I'm saying [because $g'(\xi)\approx g'(u_{j-1})$] $g(u_j)-g(u_{j-1})\approx g'(u_{j-1})(u_j-u_{j-1})$. However, using MVT to get exact equalities is quite useful. Remember, there's no use in saying $\approx$ if you can't say how big the error is. [Technical aside: we don't need the full MVT here, only the mean value inequality, which is true in generality, but I think it's clearer with the MVT] Commented Dec 3, 2023 at 12:56
• Thanks a million!! Commented Dec 3, 2023 at 12:58