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

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,

$$1=0.$$

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