Sign up ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Possible Duplicate:
Lebesgue measure on Riemann integrable function in $\mathbb{R}^2$

Is the Lebesgue integral of a positive real function of a real variable equivalent to the Lebesgue measure of the set (in $\mathbb{R}^2$) of all the points between the interval of integration and the graph of the function?

I'm asking this because all the different definitions of "length", "area", "volume", "measure" I was exposed to (Euclidean geometry, path length, measure of a set, integral, scalar product, ...) seem to be different from one another and I would like to see what are the points in common

share|cite|improve this question

marked as duplicate by Qiaochu Yuan Jun 24 '12 at 6:18

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Yes. ${}{}{}{}{}$ –  Qiaochu Yuan Jun 18 '12 at 11:57
dupe? –  leo Jun 18 '12 at 16:52

1 Answer 1

up vote 3 down vote accepted

Let $f \ge 0$ be measurable. Then (where $\chi_A$ denotes the characteristic function of a set $A$) \begin{align*} \int_{\mathbb R} f(x)\, dx &= \int_{\mathbb R} \int_{\mathbb R} \chi_{[0, f(x))}(y)\, dy\, dx\\\ \text{and by the Fubini-Tonelli theorem,} &\\ &= \int_{\mathbb R^2} \chi_{[0, f(x))}(y)\, d(x,y)\\ &= \int_{\mathbb R^2} \chi_{\{(a,b) \mid 0 \le b < f(a)\}}(x,y)\, d(x,y)\\\ &= \lambda\bigl(\{(x,y) \mid 0 \le y < f(a)\}\bigr). \end{align*}

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.