Use the monotone convergence theorem to calculate $\int_{-\infty}^{+\infty}e^{-|x|}d\bar{\lambda} $

Question

For a particular Lebesgue exercise, I am required to work out the integral $\int_{-\infty}^{+\infty}e^{-|x|}d\bar{\lambda} $ using the monotone convergence theorem. I am finding it difficult to start, mainly as the monotone convergence theorem generally involves a sequence of functions that converge to a particular function, but in this case, we are given a singular function.

What is $\overline {\lambda}$? – Kavi Rama Murthy May 01 '20 at 09:27 — Kavi Rama Murthy, May 01 '20 at 09:27
It is the Lebesgue measure on the real set R – Ann May 01 '20 at 10:45 — Ann, May 01 '20 at 10:45

score 0 · Accepted Answer · answered May 01 '20 at 11:40

0

I think what you are asked to do is write the integral as the limit of $\int I_{(-n,n)} e^{-|x|}dx$ (which is permissible by Monotone Convergence Theorem) so we get the answer as $\lim 2\int_0^{n} e^{-x}dx =\lim 2(1-e^{-n})=2$.

answered May 01 '20 at 11:40

Kavi Rama Murthy

311,013

Thank you. This makes a lot of sense – Ann May 01 '20 at 11:56
What is $I_{(-n,n)}$? – Rodrigo May 02 '20 at 00:53
@Rodrigo It is the characteristic function of the interval. – Kavi Rama Murthy May 02 '20 at 04:36

Use the monotone convergence theorem to calculate $\int_{-\infty}^{+\infty}e^{-|x|}d\bar{\lambda} $

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