Why can the limit in this case pass inside the integral?

Question

I was reading the Basel sum proof and found the following passage:

$\frac{1}{n^2}= \iint_R \,(xy)^{n-1} \,dx\,dy \quad ;\quad R=[1,0]\times [1,0] \Longrightarrow $

$ \sum_{n=1}^{\infty} \frac{1}{n^2} = \iint_R \sum_{n=1}^{\infty} (xy)^{n-1} \,dx\,dy = \iint_R \ \frac{1}{1-xy} \,dx\,dy$

I don't know why it's correct to pass the limit inside the integral. I read here on the forum that this is only possible when the integrand converges uniformly across the entire domain, which doesn't seem to be the case.

The sequence of functions $f_n = \sum_{n=1}^{\infty} x^{n-1}$ does not converge uniformly in $(0,1)$. but on any interval $(0,a) ; 0\leq a \leq 1 $

does the sequence $f_n(x,y)= \sum_{n=1}^{\infty} {(xy)}^{n-1}$ converge uniformly to $f(x,y)=\frac{1}{1-xy}$ in $[1,0] \times [1,0]$ ?

How to justify such a passage?

Answer 1

$|x^n-c^n|=|x-c|\sum_{k=0}^{n-1}x^kc^{n-1-k}<n|x-c|c^{n-1}$

$|x-c|<\frac{\epsilon}{n}$. Epsilon is independent of $c$ so we do have uniform continuity.

Generally, a bounded derivative on an interval implies uniform continuity on that interval.

Answer 2

The summands $(xy)^{n-1}$ are non-negative on $[0,1)^2$ and the partial sums $f_m(x,y) =\sum_{n=1}^m (xy)^{n-1}$ are monotonically non-decreasing with respect to $m$. Viewed either as a Lebesgue or an improper Riemann integral, it follows from the monotone convergence theorem that

$$\int_0^1\int_0^1 \frac{dx\,dy}{1-xy}= \int_0^1 \int_0^1\sum_{n=1}^\infty (xy)^{n-1}\,dx\,dy = \int_0^1\int_0^1\lim_{m \to \infty}f_m(x,y) \, dx \,dy \\ =\lim_{m \to \infty}\int_0^1\int_0^1f_m(x,y) \, dx \,dy = \lim_{m \to \infty}\sum_{n=1}^m \int_0^1 \int_0^1(xy)^{n-1}\,dx\,dy = \ldots$$

You also asked if the sequence $f_n(x,y)$ converges uniformly to $f(x,y) = \frac{1}{1-xy}$ on $[0,1)^2$. The answer is no. To show this note that

$$\sup_{(x,y)\in [0,1)^2}|f_n(x,y) - f(x,y)| = \sup_{(x,y)\in [0,1)^2}\left|\sum_{k=n+1}^\infty(xy)^{k-1}\right|\geqslant \sup_{(x,y)\in [0,1)^2}\sum_{k=n+1}^{2n}(xy)^{k-1}\\ \geqslant \sup_{(x,y)\in [0,1)^2}n \cdot (xy)^{2n}\geqslant n \left(1-\frac{1}{n}\right)^{2n}\left(1-\frac{1}{n}\right)^{2n}= n \left(1-\frac{1}{n}\right)^{4n} \underset{n \to \infty}{\longrightarrow}+\infty \cdot e^{-4}= +\infty$$

Since $\lim_{n \to \infty} \sup_{(x,y)\in [0,1)^2}|f_n(x,y) - f(x,y)| \neq 0$ the convergence is not uniform.

Uniform convergence is not a necessary condition for passing a limit inside an integral. It is also not a sufficient condition if the integral is taken over an unbounded region. A counterexample is

$$1 = \lim_{n \to \infty}\int_0^\infty \frac{1}{n} \chi_{[0,n]}(x) \, dx \neq \int_0^\infty \lim_{n \to \infty}\frac{1}{n} \chi_{[0,n]}(x) \, dx = 0$$