g_n integrable on R

Question

Let g_n (x) = 1 if x=0

           sin x /x  if -n<= x <= n
           0         if x<-n  or  x>n

show that for every n, g_n is integrable (Lebesgue integrable) on R.

Well, what's the measure of the set of discontinuity points of each $;g_n;$ ... ? — DonAntonio, Mar 17 '14 at 04:37
Have you already studied the Riemann-Lebesgue Theorem? Check this: http://math.northwestern.edu/~scanez/courses/berkeley/math104/fall11/handouts/riemann-lebesgue.pdf — DonAntonio, Mar 17 '14 at 14:49

score 0 · Answer 1 · answered Mar 17 '14 at 15:09

0

Without the R-L Theorem:

$$\int\limits_{-\infty}^\infty g_n(x)dx\stackrel ?=\int\limits_{-\infty}^{-n}g_n(x)dx+\int\limits_{-n}^ng_n(x)dx+\int\limits_n^\infty g_n(x)dx$$

Since each and every one of the three integrals on the RHS above exists, you can delete now that question mark (?) and we're done.

answered Mar 17 '14 at 15:09

DonAntonio

But I think we need to show that the integral from -n to n of g_n (x) exists?!?! – Jawad Mar 17 '14 at 15:11
Well, there $;g_n;$ is a continuous functions with a removable discontinuity point at $;x=0;$ , so no problem at all... – DonAntonio Mar 17 '14 at 15:17
But we have to show that g_n is bounded, right? – Jawad Mar 17 '14 at 15:36
Well, isn't it obvious @Jawad ? $;g_n(x);$ is non-zero in a closed interval and there it is continuous...well, less one removable discontinuity point. – DonAntonio Mar 17 '14 at 15:39
Yes. thank you. I just wanted to the last step with no doubt. Thank you!!! – Jawad Mar 17 '14 at 15:41
@DonAntonio: The boundedness is a matter of the interval being compact not closed (little but important difference). – C-star-W-star Aug 13 '14 at 09:15

1 Answers1