Sorry if this is a basic question; I'm just starting to learn measure theory.

In analysis, we have the following theorem:

Let $f:[a,b] \to \mathbb{R}$ be bounded and let $S$ denote the discontinuity set of $f$. Then $f \in \mathcal R[a,b] \Longleftrightarrow \mu(S) = 0$

I was wondering whether or not this also applies to improper integrals of the form

$$\int_a^\infty f, \int_{\mathbb{R}} f$$

For simplicity, consider only the first type of improper integral. We can have for any bounded, compact interval with $x>a$, $f:[a,x] \to \mathbb{R}$ have a discontinuity set with measure $0$. This implies

$$g(x) := \int_a^x f$$ is well defined and letting $x \to \infty$ is valid, assuming the limit exists.

However, is it ever possible for

$$\lim_{x \to \infty} g(x) = \int_a^\infty f$$

to exist, but for

$$\mu(S_{[a,\infty)}) > 0$$?

If so, how do we characterize (improper) Riemann integrals?


As part of the sentence

$\lim_{x\to\infty} g(x)$ exists

you are assuming that $g(x)$ exists for all $x > a$ (for example). In particular, you know that $f\in \mathcal{R}[a,x]$ for any $x$. This means that $S_{[a,\infty)} \cap [a,x]$ has measure zero for any $x$.

Now, since $S_{[a,\infty)} = \cup_{n\in \mathbb{Z}} \left( S_{[a,\infty)} \cap [a,n]\right)$, it follows from the countable subadditivity property of the Lebesgue measure $\mu$ that the set of discontinuous points has measure zero.

That the Lebesgue criterion is usually stated only for compact intervals is that for improper integrals this is no longer if and only if. Clearly there are bounded continuous functions $f$ defined over $\mathbb{R}$ such that the integral implied by the notation $\int_{\mathbb{R}} f ~dx$ doesn't exist.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.