What are the formal necessary and sufficient conditions for u-sub to work?

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My book made the interesting claim that, if $x=\varphi(u)$ is our u-sub function, the technique of substitution will only hold if:

  1. It is a bijection onto the domain of integration (this makes sense)
  2. It is continuously differentiable
  3. Its derivative is nowhere zero on the interior

We will be using the Riemann integral here. I am unsure if $2)$ can be weakened, or if $3$ is strictly necessary - even though I pretty much buy the book's argument, I have heard conditions $2,3$ nowhere else before so I want to check with the community if this is accurate. The book is Courant's Differential and Integral Calculus - it's good, but not 100% rigorous!

Their argument, paraprashed:

Take $I_x=[a,b]\subset\Bbb R$. Suppose there exists some $\varphi:I_u\to I_x$ which is a bijection and $I_u$ is compact. In the Riemann integral $\int_a^b f(x)\,dx$, we can substitute, taking the tagged partitions in $x,u$ as $(x_n,t_n),(u_n,s_n)$: $$\begin{align}\int_{I_x}f(x)\,dx=\sum f(t_n)(x_{n+1}-x_n)\overset{1}{=}&\sum (f\circ\varphi)(s_n)(\varphi(u_{n+1})-\varphi(u_n))\\\overset{2}{=}&\sum(f\circ\varphi)(s_n)\cdot\varphi'(\xi_n)(u_{n+1}-u_n)\\\overset{3}{\to}&\int_{I_u}(f\circ\varphi)(u)\cdot\varphi'(u)\,du\end{align}$$ Step $1$ required $\varphi$ to be a bijection; step $2$ required $\varphi$ to be differentiable on each $(u_{n+1},u_n)$, and step $3$ required $f\circ\varphi$, $\varphi'$ to both be integrable, and for $\varphi'$to be continuous as its mean value $\xi$ needs to be squeezed to the same tagged value $s_n$, by definition of Riemann integral.

Now I am aware that if I take $2,3$, the Inverse Function Theorem guarantees that $\varphi$ is indeed a bijection and that all is well. However, in the analysis of the Riemann sum, we only require that $\varphi'(\xi)$ is non-zero; the way I see it, we could still have some point inside $I_u$ where $\varphi'$ is zero, and just make a choice of partition such that this is never the mean value - this works, right? I know that if the points where $\varphi$ has $0$ derivative are maxima or minima we lose invertibility, but if it is a point of inflection then, as far as I know, we can still have it as an invertible function. Yes, its inverse would not be differentiable at that point, but we surely don't care as only $\varphi$ is involved in the integration - right?

I'm also wondering if condition $2$ can be weakened - in the Riemann integral I believe we can choose our partitions, so perhaps it would be possible to make a choice of partition $s_n$ (with $\varphi(s_n)=t_n$) such that $s_n=\xi_n$, and the continuity of $\varphi'$ could be relaxed.

What actually are the needed conditions? And if Courant is completely correct, I'd appreciate it if someone could correct my overthinking here!

Side-note: I have not studied Lebesgue integration all that much - does U-sub still work there?

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I haven't thought your thinking through completely but u-sub is essentially a special case of the (Jacobi) transformation theorem:

Let $\Omega \subseteq \Bbb R^n$ be an open set and $\phi : \Omega \to \phi(\Omega) \subseteq \Bbb R^n$ a diffeomorphism. Then $f$ is integrable on $\phi(\Omega)$ iff $x \mapsto f(\phi(x)) |\det(J_\phi(x))|$ is integrable on $\Omega$ and in this case we have $$ \int_{\phi(\Omega)}f(x) dx = \int_\Omega f(\phi(x)) |\det{J_\phi(x)}| dx $$ where $J_\phi(x)$ denotes the jacobian matrix of $\phi$.

A diffeomorphism is a bijection which is everywhere continuously differentiable with continuously differentiable inverse. Regular u-sub is the $\Bbb R$ case of this theorem.

The theorem doesn't have a standalone article on the English wikipedia, but it's listed under the integration by substitution article (https://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables) along with other generalizations and how the situation looks for the Lebesgue integral.

Taking this into account: your conclusion that $\phi'$ may be $0$ at some point is correct - this is the generalization via Sard's theorem. You may also relax the condition that $\phi'$ is continuous - but this brings with it the new condition of $\phi^{-1}$ being continuous (and having to exist).