In the Riemann theory for a function f defined on all of R, we define its improper integral over R as the sum of two limits:(adsbygoogle = window.adsbygoogle || []).push({});

[tex]\int_{-\infty}^{+\infty}f(x)dx = \lim_{a\rightarrow -\infty}\int_{a}^0f(x)dx+\lim_{b\rightarrow +\infty}\int_{0}^b f(x)dx[/tex]

and in general, this is not equal to

[tex]\lim_{L\rightarrow +\infty}\int_{-L}^L f(x)dx[/tex]

For instance, for f(x)=sin(x), neither of the two limits above exists, but the limit below is 0.

But what does Lebesgue says? We know that given f:R-->R integrable, the function [tex]\nu:\mathfrak{L}_{\mathbb{R}}\rightarrow \mathbb{R}[/tex] defined by

[tex]\nu(E)=\int_Ef(x)dx[/tex]

is a (signed) measure. Signed measures enjoy the same continuity properties as finite and positive measures. That is, if E_n is an increasing sequence of measurable sets, then

[tex]\nu\left(\bigcup_{n=1}^{+\infty}E_n\right)=\lim_{n\rightarrow +\infty}\nu(E_n)[/tex]

In particular, for E_n = ]-n,n[ and the measure defined above,

[tex]\int_{-\infty}^{+\infty}f(x)dx=\lim_{n\rightarrow +\infty}\int_{-n}^nf(x)dx[/tex]

(Since U E_n=R)

What's up with that? (as in : any comments welcome)

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# Something a little odd - Lebesgue vs Riemann integral

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