Not Lebesgue Integrable

Question

Let g(x) be the function from R to R defined by $g(x)= 1$ if $x=0$, $\frac{\sin x}x$ otherwise.

Define the function $g_n (x)= g(x)$ if $-n < x < n$ and $x=0$ otherwise.

Show that for every n, $g_n$ is integrable on $\mathbb R$. And that $g_n$ converges to $g$.

You have choices: simple convergence? uniform? $L^2$? – mookid Mar 06 '14 at 22:10 — mookid, Mar 06 '14 at 22:10

score 1 · Answer 1 · answered Mar 06 '14 at 22:19

1

$ |g_n(x) - g(x)| \neq 0 \Rightarrow |x|\ge n $ so $$ \sup_x |g_n(x) - g(x)| = \sup_{|x|\ge n} \frac{|\sin x|}{|x|} \le \frac 1n $$ then $g_n\to g$ and the convergence is uniform.

As far as the integrability is concerned, as $g_n$ is bounded and non zero on a bounded interval, it is Lebesgue integrable.

answered Mar 06 '14 at 22:19

mookid

28,236

why g_n -g not 0???? – Jawad Mar 06 '14 at 22:23
to locate where is the difference takes the biggest value. – mookid Mar 06 '14 at 22:27
since g is not integrable, why does this not violate Lebesgue's Dominated Convergence Theorem? – Jawad Mar 06 '14 at 22:28
why should this violate the TDC? – mookid Mar 06 '14 at 22:30
Because g_n goes to g but g is not integrable – Jawad Mar 06 '14 at 22:37
yes, but there is no domination hypothesis. – mookid Mar 06 '14 at 22:39
but we have |g_n| is bounded – Jawad Mar 06 '14 at 23:18
the domination hypothesis is: $\exists G \ \ |g_n|\le G$ and $\int G < \infty$. – mookid Mar 06 '14 at 23:31

Not Lebesgue Integrable

1 Answers1