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On the one hand, this question seems obviously true. However (I think that) it assumes that if two definite integrals are equal then their integrands are the same.

In particular, let us assume $f(t)=g(t)$, and define the Fourier transform such that:

$$f(t)=\int_{-\infty}^{\infty}\tilde{f}(\omega)e^{i\omega t}\frac{d\omega}{2\pi}.(1)$$

It follows that:

$$\int_{-\infty}^{\infty}\tilde{f}(\omega)e^{i\omega t}\frac{d\omega}{2\pi}=\int_{-\infty}^{\infty}\tilde{g}(\omega)e^{i\omega t}\frac{d\omega}{2\pi}.(2)$$

Does $f(t)=g(t)$ imply $\tilde{f}(\omega)=\tilde{g}(\omega)$? This seems to be obviously true, but if I look at the equation (2), it seems to require that if two definite integrals are the same, then their integrands must be equal (which I know is not true in general).

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    Of course ! If $x=y$ do you think that $f(x)\neq f(y)$. By the way, if $\hat f=\hat g$, then you also have that $f=g$ a.e. – 2017-01-30
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    The real question is whether or not $f(t) = f(t)$ implies that $\tilde{f}(t)=\tilde{f}(t)$, or rather, whether or not the Fourier transform is unique (which it is). Therefore, by uniqueness $f(t)=g(t)$ implies they must have the same transform. More interestingly, as @Surb stated above, it works the other way (except perhaps on a set of measure 0). – 2017-01-30
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    What class of functions are you considering for $\tilde{f}$ and $\tilde{g}$? – 2017-01-31

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