# Analysis Prelim prep: Lebesgue integration

1. Jul 10, 2011

### 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!

2. Jul 10, 2011

### 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 $$|F(x)-F(y)| \leq \epsilon$$.

3. Jul 11, 2011

### 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!

4. Jul 11, 2011

### disregardthat

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.)

Last edited: Jul 11, 2011