Question

question about lebesgue integration:

If I want to calculate the integral of a given function, where the function is continuous over the whole interval of integration, except it's undefined at one of the end points (division by zero), then can I take the limit approaching that end point, and use the normal Riemann integral methods?

I know there's a theorem which says that if the function is continuous on the compact interval then the lebesgue integral is the same as the Riemann integral. But is that also true when taking the limit of such integrals?

Answer 1

To a point, as long as it has a defined neighborhood, yes. This includes all points in the extended real plane.

Answer 2

if i recall correctly a function is riemann integrable if it is continuous almost everywhere

Answer 3

Thankyou.

So if I had another function where I wanted the lebesgue integral over [1,infinity), and the function is continuous over that interval, then I could just calculate it in the normal Riemann way, taking the limit?

I know there's a counter-example where a function is Riemann integrable but not Lebesque integrable. But I've forgotten how that works...

Answer 4

as far as i can remember, no one actually computes a lebasque integral, just as no one ever computes a riemann integral.

there is not function riemeann integrable that is not lebesque integrable, it is the other way around

Answer 5

that is to the best of my recollection
riemann integrable functions are all lebesque integrable. the only glitch is that the space of riemann integrable functions is not complete, i.e. you can have a sequence of riemann integrable functions that converge to a function that is no riemann integrable, whereas space of lebesque integrable functions is complete

Answer 6

My professor gave that the function f(x)=(-1)^n/n on [n,n+1), n a natural number is Riemann integrable but not Lebesgue integrable. I thought the same thing, that all Riemann integrable functions are Lebesgue integrable. But apparently there are some weird exceptions like that you can come up with?

Answer 7

and in the example above, the function is over [1,infinity). But f = (-1)^n/n for each interval [n,n+1).

Answer 8

wikipedia agrees that it's not always true: http://en.wikipedia.org/wiki/Lebesgue_integration#Limitations_of_Lebesgue_integral

This is confusing to me because I had also believed that Riemann integrable => Lebesgue integrable. But I think there must be some extra conditions on the set thats' being integrated over, or the amount of discontinuities or something.

Answer 9

If it's unbounded, it can be done with a limit extension, hence the point of doing such on the extended Real Plane. But this is not always the case for finite-difference sequences.