Most general conditions for variable substitution with Riemann integral

Question

This question is motivated by the discussion here:

https://matheducators.stackexchange.com/a/26687/117

Let $g$ be defined and differentiable on an interval containing $[a,b]$ and $f$ be defined on an interval containing the image of $g$.

What are the most general conditions on $f$ and $g$ so that the change of variables theorem holds?

$$ \int_a^b f(g(x))g'(x) \textrm{ d}x = \int_{g(a)}^{g(b)} f(u) \textrm{ d}u $$

Many textbooks list conditions like "$f$ is continuous and $g$ is continuously differentiable".

I can prove it also in the case "$f$ is Riemann integrable, $g'$ continuously differentiable and monotone."

Do we in fact have it in the more general case "$f$ is Riemann integrable, $g'$ is Riemann integrable"?

If not, I would be grateful for a counterexample. It would also be nice to have a list of various ways to loosen the conditions in various directions.

Answer 1

In the traditional formulation, the change of variables formula is

$$\int_{g(a)}^{g(b)} f(u)\,\mathrm{d}u = \int_a^b f(g(x))g'(x)\,\mathrm{d}x.\tag{1}\label{trad}$$ We typically assume that $g$ is continuous on $[a,b]$ and differentiable on $(a,b)$. With the additional assumption that $g'$ is Riemann integrable on $[a,b]$, the fundamental theorem of calculus guarantees that $g$ is an indefinite integral of $g'$, so we have the following broader formulation.

Let $g$ be Riemann integrable on $[a,b]$, and let $G$ be an indefinite integral of $g$ on $[a,b]$, i.e. $G(t)=c+\int_a^t g(x)\,\mathrm{d}x$ for some $c\in\mathbb{R}$. The change of variables formula is then

$$\int_{G(a)}^{G(b)} f(u)\,\mathrm{d}u = \int_a^b f(G(x))g(x)\,\mathrm{d}x.\tag{2}\label{formula}$$

Kestelman (1961) is attributed with the proof of \eqref{formula} when $f$ is Riemann integrable on $G([a,b])$.

The other direction is attributed to Preiss and Uher (1970). If $f$ is bounded on $G([a,b])$ and $(f \circ G)g$ is Riemann integrable on $[a,b]$, then \eqref{formula} holds.

Sarkhel and Výborný (1996/1997) and Tandra (2015) present alternative proofs of equivalent formulations. Tandra (2015) replaces the assumption that $G$ is an indefinite integral of $g$ with the assumption that $G$ is Lipschitz continuous and $G'=g$ almost everywhere, in the spirit of the traditional formulation.

Kuleshov (2021) weakens the condition in Preiss and Uher's result that $f$ must be bounded on $G([a,b])$ to boundedness of $f$ on $[G(a),G(b)]$.

Answer 2

The theorem holds - for a Riemann integral - when both sides are defined and when $f$ is antidifferentiable. If $g$ is everywhere differentiable and $(f\circ g)\cdot g'$ and $f$ are both Riemann integrable then the following Riemann integrals exist (by definition) and agree: $$\int_a^b(f\circ g)g'=\int_{g(a)}^{g(b)}f$$

The proof is simple; let $F$ be any primitive for $f$. $F\circ g$ is a primitive for $(f\circ g)g'$ so by one of the fundamental theorems of calculus we get the LHS to (exist and equal) $F(g(b))-F(g(a))$ which is precisely the RHS (by the other fundamental theorem of calculus).

Note $(1)$: We don't always use $u$-substitution quite in the way given by the above. If we recognise our integrand as of the form $(f\circ g)g'$ then by "letting $u=g(x)$", all is well. But, it is very common to instead have an integrand just of the form "$f$" and we let $u=g(x)$... and actually we try to invert this. Stronger hypotheses are needed in that case, namely that $g$ is invertible and its inverse is (i) differentiable (ii) $(f\circ g^{-1})(g^{-1})'$ is Riemann integrable. Things can easily go wrong, for instance it is not so uncommon for students to do something like this:

$$\frac{2}{3}=\int_{-1}^1x^2\,\mathrm{d}x\overset{u=x^2}{\underset{???}{=}}\int_0^1\frac{1}{2}\sqrt{u}\,\mathrm{d}u=\frac{1}{3}$$

Note $(2)$: to extend this in general to Lebesgue integration (taking $f$ to be Lebesgue but not Riemann integrable, e.g.) is subtle. There is a change-of-variables formula similar to this: $$\int_{[a,b]}(f\circ g)|g'|=\int_{g([a,b])}f$$In the instance that $g$ is $C^1$ with nonvanishing derivative and $f$ is in $L^1$. Note the absolute value $|g'|$; Lebesgue integrals don't have the same orientation concerns that Riemann integrals do. However there are also Lebesgue-integral forms of the classical FTC (see Rudin):

$$f(b)-f(a)=\int_{[a,b]}f'$$Either when: (i) $f$ is absolutely continuous (thus $f'$ exists a.e.) and (ii) $f'$ is in $L^1[a,b]$ or when: (i) $f$ is differentiable everywhere and (ii) $f'$ is in $L^1[a,b]$

Working with this, we recover the Riemann change of variables formula: $$\int_a^b(f\circ g)g'=\int_{g(a)}^{g(b)}f$$If in your situation we have that $f$ and $(f\circ g)g'$ are both in $L^1$ and either: (i) $g$ is absolutely continuous and $f$ has an absolutely continuous (a.e.)-antiderivative or (ii): $f$ has an (everywhere-)antiderivative and $g$ is differentiable everywhere.