Simplify Lebesgue intergral to become Riemann Integral

Asked Sep 26 '17 at 15:03

Active Sep 26 '17 at 15:30

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I have reviewed some Lebesgue measure and Lebesgue integral, and I have a quick question:

Given $x\in R^2$ with Lebesgue integral $\int_{A} e^{-x^2}du(x)$, where $u(\cdot)$ is some Lebesgue measure and A is any subsets of $\mathscr{B}(R^2)$. If we choose $A = R^2$ (essentially, the whole xy-plane), could we write $\int_{A} e^{-x^2}du(x)= \int_{-\infty}^{\infty} e^{-x^2}dx$, which is essentially a Riemann integral? If not, do we have any ways to get rid of $du(x)$ without going through the route of using simple functions?

edited Sep 26 '17 at 15:30

asked Sep 26 '17 at 15:03

ghjk

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What do you mean by "a" Lebesgue measure? Also, wherever well defined, the Riemann integral is the same as the Lebesgue integral. – Sarvesh Ravichandran Iyer Sep 26 '17 at 15:11
well, a Lebesgue measure $u$ in this case is $u([a,b]) = b-a$? And where did you get the statement: "wherever well defined, the Riemann integral is the same as the Lebesgue integral"? I understand Lebsgue integral is an extension of Riemann integral for the extreme cases, but your statement is quite strong. – ghjk Sep 26 '17 at 15:19
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This page may be of help : https://math.stackexchange.com/questions/1065151/if-a-function-is-riemann-integrable-then-it-is-lebesgue-integrable-and-2-integr – Sarvesh Ravichandran Iyer Sep 26 '17 at 15:30
You were right. I fixed the definition of the Lebesgue measure in this case. Could you please take a look at the OP again? – ghjk Sep 26 '17 at 15:31
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Yes, I think you can write the statement you have written. – Sarvesh Ravichandran Iyer Sep 26 '17 at 15:36
Many thanks for your help, Sir! Is it due to the same reason in your link, even when the domain is 2-dimensional? – ghjk Sep 26 '17 at 15:39
Yes, you are correct. – Sarvesh Ravichandran Iyer Sep 26 '17 at 15:40

Simplify Lebesgue intergral to become Riemann Integral

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