Analysis Prelim prep: Lebesgue integration

[laonious]

Hi everyone,
I am studying past analysis prelim exams to take in the fall and have run into one which really has me stumped:

Let f be a real-valued Lebesgue integral function on [0,\infty).
Define
F(x)=\int_{0}^{\infty}f(t)\cos(xt)\,dt.
Show that F is defined on R and is continuous on R.
Show that \lim_{\rightarrow \infty}F(x)=0.

That F is defined is straightforward I think, the part I am struggling with is showing that it is continuous.

Going the route of a direct proof:
F(x)-F(y)$=\int_{0}^{\infty}f(t)(\cos(xt)-\cos(yt))\,dt,
but it isn't clear to me how to show that this is enough to make the integral small.
Any ideas would be greatly appreciated, thanks!

 

[disregardthat]

|\cos(xt)-\cos(yt)| <= |t||x-y||\sin(ct)| <= |t||x-y| where we use the mean value theorem.

Also, |cos(xt)-cos(yt)| <= 2 for all x,y and t.

Thus |\int^{\infty}_0f(t)(\cos(xt)-\cos(yt)) dt| \leq \int^{\infty}_0|f(t)||\cos(xt)-\cos(yt)| dt \leq \int^{r}_0|f(t)||t||x-y| dt + \int^{\infty}_r2|f(t)| dt

Now, pick an epsilon. Let r be a real number such that \int^{\infty}_r2|f(t)| dt \leq \frac{\epsilon}{2}. We have that

\int^{r}_0|f(t)||t||x-y| dt \leq |x-y|r\int^{r}_0|f(t)| dt, so let \delta = \frac{\epsilon}{2r\int^{r}_0|f(t)| dt}. Thus |F(x)-F(y)| \leq \epsilon.

 

[laonious]

Wow, that's clever breaking it up into a finite and an improper integral. I'd tried using the mean value theorem but wasn't sure how to show integrability. Thanks for your help!

 

[disregardthat]

Wow, that's clever breaking it up into a finite and an improper integral. I'd tried using the mean value theorem but wasn't sure how to show integrability. Thanks for your help!

No problem. Note though we are not transitioning into improper and finite riemann-integrals if that was what you were implying; the conditions of f does not imply riemann-integrability. Think of it as applying the indicator function wrt the area over which it is integrated in my "riemann integral notation", something I suppose the problem itself assumed by the wording of it.

(That such an r exists is not immediately obvious, we use that u(A) = int_A 2|f(t)| dt (lebesgue-integral) is a measure , and that lim_(A \to emptyset) u(A) = 0. It can perhaps be proven more easily.)