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I am trying to understand how "the" general integral is defined in measure theory but I just don't get it. I'm using Friedman's "Foundations of Modern Analysis".

A simple function $g=\sum \beta_i 1_{E_i}$ is integrable if $\mu(E_i)<\infty$ for all $i$, and the integral of $g$ is $\int g \space d\mu := \sum \beta_i \mu(E_i)$. If $E$ is a measurable set and $f$ an integrable simple function, the integral of $f$ over $E$ is $\int_E f \space d\mu := \int f \space 1_E \space d\mu$. Since $f$ is simple, it can be written as $f=\sum \alpha_i 1_{E_i}$ (for what $E_i$?). We then have $\int_E f \space d\mu= \int 1_E (\sum \alpha_i 1_{E_i})d\mu$. The presence of both $1_E$ and the $1_{E_i}$ causes me some trouble but I think the integral comes out as $\sum \alpha_i \mu(E \cap E_i)$. This is quite constructive.

Now then, a measurable function $f:\Box \rightarrow \bar{\mathbb{R}}$ is said to be integrable if there exists a sequence $\{f_n\}$ of simple integrable function such that $\{f_n\}$ is a Cauchy sequence in the mean and $\mathrm{lim} f_n(x)=f(x)$ a.e. (or the sequence converges in measure to $f$).

This is not quite as constructive. Forgetting the high school stuff for a moment, none of this really tells me how to integrate even a constant function, how would I do that with the above information? Or is that focusing on the wrong thing? Is the important thing, from a course as this (I assume first courses in measure theory are the same around the world), mainly to learn the qualitative aspects of measure theory and how it encompasses Riemann integration etc, rather than knowing how to do actual integration?

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A constant function is a simple function, so if you can handle those, you're done with constant functions. The integral of a constant function is the constant value times the measure of the domain. –  Michael Hardy Nov 26 '13 at 17:54

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The integral of a non-negative function $f$ is defined as the smallest number that's not too small to be the integral, and is equal to $\infty$ if all numbers are too small.

A number is too small to be the integral if it is less than the integral of some simple function that is everywhere less than or equal to $f$. (A simple function is one that takes only finitely many values.)

For non-positive-valued functions $f$, find the integral of $-f$ according to the above and then multiply by $-1$.

For functions taking both negative and non-negative values, treat the negative and non-negative parts separately and then add. The case of $\infty-\infty$ is left undefined.

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Thanks for your answer, but I'm not sure how this addresses my question(s)? :) –  Erik Vesterlund Nov 26 '13 at 22:34

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