1
$\begingroup$

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?

$\endgroup$
  • 1
    $\begingroup$ If $f$ is $L^1$ then every integral here is a Lebesgue integral. $\endgroup$ – Angina Seng Apr 7 at 18:26
  • $\begingroup$ @AnginaSeng My book specifies that notation($\int_{-\infty}^{\infty}$) as a Riemann integral as opposed to Lebesgue($\int_{\mathbb{R}}$) so I assumed the integral in the question is also Riemann, though I am not 100% sure either. $\endgroup$ – DayDream Apr 7 at 18:30
  • $\begingroup$ As mentioned above, if $f$ is only assumed $L^1$ then every integral here needs to be interpreted as a Lebesgue integral. Consider the typical example of $f$ being the indicator function of the rationals (or the cantor set) or any other "crazy" $L^1$ function. These are not Riemann integrable (not even in the proper sense with finite bounds) so there's no meaning to the improper integrals if you consider everything as Riemann integrals. If you wish to consider everything in the Riemann sense, you need to impose more regularity on $f$, for example being continuous and decaying fast enough $\endgroup$ – peek-a-boo Apr 7 at 18:48
1
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.