What's wrong with this proof that $0 = 1$?

Question

Let $$f_n(x)=\frac{1}{\sqrt{\pi n}}e^{-x^2/n}.$$ Note that $f_n(x)\to 0$ uniformly as $n\to\infty$. [Proof: $0\leq f_n(x)\leq\frac{1}{\sqrt{\pi n}}$; given any $\epsilon > 0$, let $M=\left\lceil\frac{1}{\pi\epsilon^2}\right\rceil$. This guarantees that $\forall n>M:\forall x:|f_n(x)-0|<\epsilon$.]

Uniform convergence justifies taking the limit $n\to\infty$ under the integral sign:

$$\lim_{n\to\infty}\int_{-\infty}^{+\infty} f_n(x)\,dx = \int_{-\infty}^{+\infty} \lim_{n\to\infty} f_n(x)\,dx$$

The left-hand side is $1$, because

$$\forall n>0:\int_{-\infty}^{+\infty} f_n(x)\,dx = 1,$$

whereas the right-hand side is $0$, because

$$\forall x\in\mathbb{R}:\lim_{n\to\infty} f_n(x) = 0.$$

Therefore,

$$1=0.$$

I think you need uniform convergence on a closed interval: http://www.math.drexel.edu/~tolya/limit%20of%20integrals.pdf — Ethan Bolker, Apr 21 '19 at 23:04
@Clayton Was mistake. I retracted the flag within like 30 seconds. Sorry — user106860, Apr 21 '19 at 23:05
This question exactly shows how important it is to understand the difference between Riemann integrals and improper integrals. — Mark, Apr 21 '19 at 23:06

score 12 · Accepted Answer · answered Apr 21 '19 at 23:03

12

Uniform convergence justifies taking the limit under the integral sign for functions with bounded domain, not for functions whose domain is $\mathbb R$.

answered Apr 21 '19 at 23:03

José Carlos Santos

427,504

score 4 · Answer 2 · answered Apr 21 '19 at 23:05

Uniform convergence does not justify the interchange for an integral over an infinite interval. As an example, take $f_n(x) =1/n$ for $0\leqslant x\leqslant n$ and $f_n(x) =0$ for $x>n$.

If the improper integral is also uniformly convergent then it is permissible.

What's wrong with this proof that $0 = 1$?

2 Answers2