# When can the order of limit and integral be exchanged?

1. I was wondering for a real-valued function with two real variables, if there are some theorems/conclusions that can be used to decide the exchangeability of the order of taking limit wrt one variable and taking integral (Riemann integral, or even more generally Lebesgue integral ) wrt another variable, like $$\lim_{y\rightarrow a} \int_A f(x,y) \, dx = \int_A \lim_{y\rightarrow a} f(x,y) \,dx \text{ ?}$$
2. If $y$ approaches $a$ as a countable sequence $\{y_n, n\in \mathbb{N}\}$, is the order exchangeable when $f(x,y_n), n \in \mathbb{N}$ is uniformly convergent in some subset for $x$ and $y$?
3. How shall one tell if the limit and integral can be exchanged in the following examples? If not, how would you compute the values of the integrals:

• $$\lim_{y\rightarrow 3} \int_1^2 x^y \, dx$$
• $$\lim_{y\rightarrow \infty} \int_1^2 \frac{e^{-xy}}{x} \, dx$$

Thanks and regards!

## 2 Answers

The most useful results are the Lebesgue dominated convergence and monotone convergence theorems.

• @Robert: Thanks! (1) For both theorems you mentioned, they apply to a discrete sequence of functions. In my questions, the index is continuous. How would that be coped with? (2) Does exchangeability of limit and differentation/integral for a sequence of uniform convergent functions also apply to Lebesgue integral? – Tim Aug 24 '11 at 1:21

@Tim: You wrote in a comment: "For both theorems you mentioned, they apply to a discrete sequence of functions. In my questions, the index is continuous. How would that be coped with?"

If $\lim_{y\to a} f(x,y)$ exists, then $\lim_{n\to\infty} f(x,y_n)$ exists, for every sequence $\{y_n\}_{n=1}^\infty$ that approaches $y$, and conversely. You can use that to show that the dominated convergence theorem and the monotone convergence theorem still work in the "continuous" setting.

• Thanks! (1) Why is that "conversely"? (2) I also wonder Does exchangeability of limit and integral for a sequence of uniform convergent functions apply not only to Riemann integral but also to Lebesgue integral? – Tim Aug 24 '11 at 4:11
• @Tim: suppose $\lim_{y \to a} f(x,y)$ doesn't exist. This tells you that there is an $\epsilon$ such that for every $\delta$ there are infinitely many numbers $z$ such that $| z - y| < \delta$ but $|f(x,z) - f(x,y)| > \epsilon$. Now (and this relies on some kind of axiom of choice) you can construct a non-convergent sequence by picking one of the $z$ like above for every $\delta = 1/n$. Obviously this sequence must go to $y$ but it will stay bounded away from $f(x,y)$ by at least $\epsilon$. – Marek Aug 24 '11 at 6:20
• @Marek: Assuming $f$ is continuous, Axiom of Choice is not needed. You could e.g. take $z_n$ to be the least $z$ in $[a-1/n, a+1/n]$ such that $|f(x,z) - f(x,a)| \ge \epsilon$. – Robert Israel Aug 24 '11 at 6:49
• @Tim: (2) Yes. The convergence theorems are actually better for Lebesgue than for Riemann integrals. For example, the pointwise limit of a sequence of bounded Riemann integrable functions might not be Riemann integrable, but it is Lebesgue integrable. This problem doesn't arise for uniform convergence, though. – Robert Israel Aug 24 '11 at 6:58
• @Robert: naturally but that's a rather strong requirement on $f$. Of course, for the purpose of this question one does not need the full generality of the equivalence of two limit definitions, so good point, I guess. – Marek Aug 24 '11 at 7:39