Let $h(x_1,x_2)$ be a function of two variables from $L^2$, such that it's Fourier transform exists and is given by:
\begin{equation} \hat{h}(f_1, f_2) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}dx_1 dx_2 e^{-2\pi i (x_1 f_1 + x_2 f_2)} h(x_1, x_2) \end{equation}
And let:
\begin{equation} H(x_1, x_2) = \begin{cases} 0, & \text{if $x_1 = x_2$}\\ h(x_1, x_2), & \text{else} \end{cases} \end{equation}
What would be the Fourier transform of $H$ and can it be expressed with $\hat{h}(f_1, f_2)$?
\begin{equation} \hat{H}(f_1, f_2) = \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}dx_1 dx_2 e^{-2\pi i (x_1 f_1 + x_2 f_2)} H(x_1, x_2) \end{equation}
I am aware that, if the integrals above are interpreted as Lebesgue integrals, then
\begin{equation} \hat{H}(f_1, f_2) = \hat{h}(f_1, f_2) \end{equation}
since $h$ and $H$ differ only at points lying on the $x_1 = x_2$, which is a set with Lebesgue measure 0 in $\mathbb{R}^2$.
On the other hand, I am using $H$ in combination with $\delta$-distributions, where clearly:
\begin{equation} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}dx_1 dx_2 h(x_1, x_2)\delta(x_1 - x') \delta(x_2 - x') = h(x', x') \end{equation}
\begin{equation} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty}dx_1 dx_2 H(x_1, x_2)\delta(x_1 - x') \delta(x_2 - x') = 0 \end{equation}
The property that $\int_{-\infty} ^\infty f(x) \delta(x-x_0) dx = f(x_0)$ holds only when $f$ is continuous at $x_0$. Adapting this to your question, if $h$ is continuous at $(x', x')$ (or perhaps we only need it to be continuous in each argument but not together?) then $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}dx_1 dx_2 h(x_1, x_2)\delta(x_1 - x') \delta(x_2 - x') = h(x', x')$ is true, but $\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}dx_1 dx_2 H(x_1, x_2)\delta(x_1 - x') \delta(x_2 - x') = 0$ is false (unless, of course, $H(x', x') = 0$).