For $f\in L^1(\mathbb{R})$, show that $\lim_{\varepsilon \to 0}\int_{-\infty}^\infty \cos(\varepsilon x)f(x) \, dx=\int_{-\infty}^\infty f(x)\,dx$ Question: For $f\in L^1(\mathbb{R})$, show that
$$\lim_{\varepsilon \rightarrow 0}\int_{-\infty}^\infty \cos(\varepsilon x)f(x)\,dx=\int_{-\infty}^\infty f(x)\,dx $$
where the integral is the Riemann integral. End Question
I first thought this was pretty easy, using the dominated convergence theorem for
$$f_n(x) = \cos\left(\frac{1}{2^n}x\right)f(x),$$
but I realized I have to first change the integral to Lebesgue integral then change the order of limit
, i.e.
$$\lim_{n\to \infty}\lim_{A\to \infty}\int^A_{-A}f_n(x) \, dx = \lim_{A\to \infty}\lim_{n\to \infty}\int^A_{-A}f_n(x) \, dx$$
since I only proved that Riemann = Lebesgue holds for closed interval and bounded $f$, and
the dominated convergence theorem works for the Lebesgue integral.
Can I easily change the order of limits here? Is there any general rule for changing the order of limits?

Crucially, for an $L^1$ function, most of the mass will be concentrated on a large interval and it is this same large interval where each integrand in your sequence will have most of their mass. 
