Prove convergence of improper integral using change of variable.

Question

This may be trivial, but I could use some help...

Consider a real function $f: (0,1) \rightarrow \mathbb{R}$, continuous, positive, but not necessarily bounded. Let $g: [0,1] \rightarrow [0,1]$ be a diffeomorphism such that $g(0)=0$, $g(1)=1$ and the integral

$$ I = \int_{0}^{1} f(g(x))|g^{\prime}(x)|\,dx $$

converges.

Is it immediate that $$ \int_{0}^{1} f(x)\,dx $$ exists and equals $I$?

Is there any difference if we use Riemann or Lebesgue integration?

Answer 1

It is not immediate if you start from scratch: the change of variables formula is not a triviality even in one dimension. But it's true for both Riemann and Lebesgue integrals. If you already know the result for bounded functions, then truncation gives it in general. The form of truncation may differ: for the Riemann integral it is more natural to consider $\int_\epsilon^{1-\epsilon}$ and then let $\epsilon\to 0$; for the Lebesgue integral I would let $f_M = \min(f,M)$ and then let $M\to \infty$.

Observe that the absolute value is not needed for $g'$: it's automatically positive under the assumptions in your post. But if you also want to consider orientation-reversing diffeomorphisms ($ g(0)=1$ and $g(1)=0$) then you should watch out for the signs.

• The Riemann integral is secretly an integral of a differential form over an oriented $1$-simplex. Hence, $\int_1^0 f(x)\,dx = -\int_0^1 f(x)\,dx$, and if $g$ is in the game, you should not take the absolute value of $g'$.
• The Lebesgue integral over $[0,1]$ is usually understood as an integral of a scalar function over a measure space. It does not involve orientation. One may want to emphasize this fact by writing $\int_{[0,1]} f(x)\,dx$ or $\int_0^1 f(x)\,|dx|$.

Of course, one can (and should) use the Lebesgue integral as the basis for the development of differential forms, but that usually comes later. When working with $[0,1]$ in the context of Lebesgue theory, people normally treat it just as a measure space.