In Rudin's Real and Complex Analysis, there is the following result about Fourier transforms.
The Uniqueness Theorem If $f\in L^1(\mathbb{R})$ and $\hat{f}(t)=0$ for all $t\in\mathbb{R}$, then $f(x)=0$ almost everywhere.
Isn't the assumption "for all $t\in\mathbb{R}$" unnecessarily too strong? I am pretty sure that we only need that $\hat{f}(t)=0$ almost everywhere to conclude that $f(x)=0$ almost everywhere. But Rudin is a very smart guy, so I guess there is a good reason for saying "for all $t\in\mathbb{R}$".
Note: In Rudin, the Uniqueness Theorem is a corollary of the following theorem.
The Inversion Theorem If $f\in L^1(\mathbb{R})$ and $\hat{f}\in L^1(\mathbb{R})$, and if $$g(x)=\frac{1}{\sqrt{2\pi}}\int_{-\infty}^\infty \hat{f}(t)e^{ixt}dt\qquad(x\in\mathbb{R}),$$ then $g\in C_0$ and $f(x)=g(x)$ almost everywhere.
As Ian pointed out, the Fourier transform of an $L^1$ function is continuous. For continuous functions, being zero a.e. and being zero everywhere are equivalent; so it makes more sense to use the shorter version, without a.e.