Lebesgue integral over the Riemann integral

[symbol0]

You always see in books that one advantage of the Lebesgue integral over the Riemann integral is that a sequence of continuous functions f_n does not have to converge unifomly to a function f to have:
integral of the limit of the sequence = the limit of the integrals of functions in the sequence

Can anybody give me an example where this equality fails using the Riemann integral, and does not fail using the Lebesgue integral?

Thank you

 

[quasar987]

The advantage of the Lebesgue integral over the Riemann integral concerning the switching of the limit and integral sign is that for Riemann, the only theorem we have is that for the switching to be justified, the sequence of function must converge uniformly. But with the Lebesgue point of view, we have also the monotone convergence theorem and the dominated convergence theorem, which say that the switchiong is justified under very mild assumptions.

Note that you will not find an example where the switching fails using the Riemann integral, and does not fail using the Lebesgue integral because if the Riemann integral exists, then the Lebesgue integral exists and is equal to it.

 

[symbol0]

Thanks quasar,
I don't think I understand completely.

Are you saying that the sequences for which the equality works/fails using the Lebesgue integral are the same sequences for which the equality works/fails correspondingly, using the Riemann integral?

If that's the case, can you give me a specific example where I can see how the Lebesgue integral gives me an advantage over the Riemann integral (regarding the switch of the limit and the integral).

 

[quasar987]

Thanks quasar,
I don't think I understand completely.

Are you saying that the sequences for which the equality works/fails using the Lebesgue integral are the same sequences for which the equality works/fails correspondingly, using the Riemann integral?

Granted that we are talking about sequences of Riemann-integrable functions then yes.

If that's the case, can you give me a specific example where I can see how the Lebesgue integral gives me an advantage over the Riemann integral (regarding the switch of the limit and the integral).

Try to prove that given any sequence of positive,increasing and converging Riemann-integrable functions {f_n}, we have

\lim_n\int_a^b f_n =\int_a^b \lim_nf_n

Or that given any sequence of converging Riemann-integrable functions {f_n} with |f_n|<g for some integrable function g, we have

\lim_n\int_a^b f_n =\int_a^b \lim_nf_n

From what I understand, people did not know of these results before Lebesgue's thesis. And it is them that drew attention to Lebesgue's construction as a worthy alternative to Riemann's definition of the integral.

 

[symbol0]

So then I guess that the answer to my original question (the example) is that the switch fails with the Riemann integral and does not fail with the Lebesgue integral for all those sequences of Riemann integrable functions where the limit of the sequence is not Riemann integrable, but it is Lebesgue integrable.

Say for instance a sequence of bounded continuous functions on [0,1] that converges to
f(x)= 0 if x is rational and 1 if x is irrational.
Then the switch would fail with the Riemann integral because f is not Riemann integrable, but the switch would not fail with the Lebesgue integral.

Right?

 

[quasar987]

So then I guess that the answer to my original question (the example) is that the switch fails with the Riemann integral and does not fail with the Lebesgue integral for all those sequences of Riemann integrable functions where the limit of the sequence is not Riemann integrable, but it is Lebesgue integrable.

Well, if you insist, but it seems weird to say that the switching fails in this case since the notion of integral for the limit function does not even make sense. A real example of the switching failing would be if all the \int f_n exist and converge and \int \lim_nf_n exists but \lim_n \int f_n \neq\int \lim_nf_n

Say for instance a sequence of bounded continuous functions on [0,1] that converges to
f(x)= 0 if x is rational and 1 if x is irrational.
Then the switch would fail with the Riemann integral because f is not Riemann integrable, but the switch would not fail with the Lebesgue integral.

Right?

I don't know if there is such a sequence of continuous functions, but if you let {r_n} denote a sequence that takes all the rational values in [0,1] and then set f_n(x)=1 if x=r_i (i=1,...,n) and 0 otherwise, then all the f_n are integrable but the limit is your function f.

 

[symbol0]

Thanks quasar