# Not riemann integrable => not lebesgue integrable ?

your post was extremely enlightening, thank you. although i kind of lost jarle and you in posts following :-)
(Concerning the Jarle-discussion: ignore it, it turned out he had misread my post, so no worries)

I'm very happy it was enlightening! :)

disregardthat

your post was extremely enlightening, thank you. although i kind of lost jarle and you in posts following :-)
I am sorry, as mr.vodka says I misinterpreted his post. I thought he was talking of a previous example provided. I should have read his post more carefully, but it's cleared up now.

lavinia
Gold Member

Hi,

In general if a function is not Riemann integrable does this mean the function is also not Lebesgue integrable? Why or why not?

I know that if the the function is Riemann integrable then its Lebesgue integrable, but I can't find any info about non-Riemann integrable functions. Maybe I'm not thinking about this in the right way...

For example, if you wanted to show 1/x is not Lebesgue integrable it would be very convenient if all you needed to show was that the Riemann integral was infinite. Otherwise I'm not sure how to compute the Lebesgue integral...Is there a straightforward method to computing integrals of such simple functions?

Any thoughts would be appreciated!
The Lebesque integral is more general than the Riemann integral. A Riemann integrable function must be continuous almost everywhere. A Lebesque integrable function need not be continuous anywhere. So the characteristic function of the Cantor set is Riemann integrable because the Cantor set has measure zero but the characteristic function of the rationals is not Riemann integrable because it is discontinuous at every point.

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Landau

These two statements are contradictory, because the defined f(x) is continuous almost everywhere.
Indeed. So given that I was right, we are forced to conclude that AlephZero's was wrong. :tongue: Indeed, his function is Riemann-integrable with integral equal to 0.

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OP, I don't know anything about lebesgue integration besides that its more general.

So, considering that I would say its safe to say that lebesgue integration works on a larger class of functions, so the answer is no.

mathwonk