For every $f \in L^1(\mathbb{R^n})$ let
$$ \hat{f} : \mathbb{R}^n \to \mathbb{C}, \quad \hat{f}(\xi) := \frac{1}{(2\pi)^{n/2}} \int_{\mathbb{R}^n} f(x) \exp{(-i \langle x, \xi \rangle)} \;\mathrm{d}x $$
be the Fourier Transform of $f$ with respect to the Lebesgue-measure.
One can show, that $\hat{f} \in L^\infty$ from the following inequality $$ \tag{$\ast$} \| \hat f \|_\infty \leq \frac{1}{(2\pi)^{n/2}} \; \|f\|_1. $$
Another theorem states that for $f,g \in L_1(\mathbb{R^n})$ the functions $\hat{f}g$ and $f \hat{g}$ are integrable with
$$ \int_{\mathbb{R}^n} \hat{f}(x)g(x) \;\mathrm{d}x = \int_{\mathbb{R}^n} f(y)\hat{g}(y) \;\mathrm{d}y. $$
The proof makes use of Fubini's theorem, therefore the integrability of $\hat{f}g$ and $f\hat{g}$ is crucial. My coursenotes say that integrability follows from ($\ast$), but I am not sure how to do this. Here is my attempt at a proof:
I need to show that $\| \hat{f}g \|_1 < \infty$. One could calculate
$$ \| \hat{f}g \|_1 = \int_{\mathbb{R}^n} | \hat{f} g | \; \mathrm{d}x \leq \|\hat{f}\|_{\infty} \|g \|_1 < \infty $$
and use the fact that $\hat{f} \in L_\infty$, but am I correctly using the monotonicity of the Lebesgue-Integral? $|\hat{f}| \leq \|\hat{f}\|_\infty$ is only true almost everywhere.
Yes, you've used it correctly.
It's true that $|\hat{f}| \le \|f\|_{L^\infty}$ only almost everywhere, but that is sufficient for your conclusion. To see why, let $A = \{|\hat{f}| \le \|f\|_{L^\infty}\}$, and break your integral into two integrals, over the sets $A$ and $A^c$. Then note that $A^c$ has measure zero.
Intuitively, the Lebesgue integral ignores null sets, so for the purposes of integration, you should expect that an "almost everywhere" statement is just as good as "everywhere".