Updated: Clarified the question.
I'm stuck in what must be a basic misunderstanding of the necessary hypotheses for the Poisson Summation Formula.
Suppose $f(x) = \frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}$, then we have
$$\hat{f}(\omega) = \int_{-\infty}^\infty f(x)e^{-2\pi i x w} \, dx = e^{-2\pi^2 \omega^2}.$$
The source I'm using states the Poisson Summation Formula as follows:
Suppose $L > 0$ Let $f \in L^1(\mathbb{R})\cap L^2(\mathbb{R})$ with $\hat{f} \in L^1(\mathbb{R})$. Further suppose that $$\sum_{k=-\infty}^\infty \hat{f}\left(\frac{k}{L}\right) < +\infty,$$ then $$ \sum_{k = -\infty}^\infty f(kL) = \frac{1}{L} \sum_{k=-\infty}^\infty \hat{f}\left( \frac{k}{L} \right ).$$
It seems to me that this can't be correct as stated, since for any $f$ satisfying the hypothesis, we can change the function on a set of measure zero and still satisfy the hypothesis, but change the left hand side of the formula without changing the right.
For example, if $f(x) = \frac{1}{\sqrt{2\pi}}e^{\frac{-x^2}{2}}$, then we have
$$\hat{f}(\omega) = \int_{-\infty}^\infty f(x)e^{-2\pi i x w} \, dx = e^{-2\pi^2 \omega^2}.$$
If we let $$g(x) = \begin{cases} 0 & x = kL, k \in \mathbb{Z} \\ f(x) & \text{else} \end{cases}$$
Then, since $g = f$ a.e., we have $\hat{g} = \hat{f}$. But this would imply $$\sum_{k = -\infty}^\infty f(kL) = \frac{1}{L} \sum_{k=-\infty}^\infty \hat{f}\left( \frac{k}{L} \right ) = \frac{1}{L} \sum_{k=-\infty}^\infty \hat{g}\left( \frac{k}{L} \right ) = \sum_{k=-\infty}^\infty g(kL) = 0$$ which is wrong.
What's wrong with these hypothesis? Can anyone provide a bit of guidance on what's the correct way to think about necessary conditions for the PSF?