Let $f(x) = \sin(x)/x$. Consider the following:
$$\int_{\mathbb{R}} f(x)dx = \lim_{n\to\infty} \int_{\mathbb{R}} f(x)\chi_{[-n,n]}dx$$
so that each $\int_{\mathbb{R}} f(x)\chi_{[-n,n]}dx$ is Riemann integrable and so it is Lebesgue integrable (integrating over compact space). Since $\int_{\mathbb{R}} f(x)\chi_{[-n,n]}dx \to \int_{\mathbb{R}} f(x)dx$ and so by completeness of $L^1(\mathbb{R})$, we have that $f\in L^1(\mathbb{R})$ so in particular, $\int_{\mathbb{R}} f(x)dx < \infty$ in the Lebesgue integral sense.
I know I have made a mistake here somewhere but I am having trouble seeing exactly where. Any help is appreciated.
$L^1(\mathbb R)$ is complete with respect to the $L^1(\mathbb R)$ norm. You would need to show that $f\cdot\chi_{[−n,n]}$ converges to $f$ in the $L^1$ norm and not just pointwisely. In fact, $\|f\cdot\chi_{[−n,n]}\|_1$ is not even bounded, so the sequence does not converge in $L^1$. It is easier to see the error if you just consider the sequence $\chi_{[-n,n]}$. Each function is $L^1$, but it does not converge with respect to the norm and the pointwise limit of the functions is clearly not in $L^1$.