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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $[0,\infty]$ be equipped with the order topology. (That is, it is a subspace of the standard topology on the extended real)

Let $(X,\mathfrak{M},\mu)$ be a measure space.

Let $f:X\rightarrow [0,\infty]$ be a measurable function, which means that the inverse of each borel set of $[0,\infty]$ under $f$ is an element of $\mathfrak{M}$.

In measure theory, we define it's integration as the supremum of $\sum_{x\in s(X)} x\mu(s^{-1}(x))$ where $s$ is a simple measurable fuction such that $s≦f$.

The point is, the role of $\mu$ only acts on those simple functions $s$, not $f$ directly.

What would go wrong if one defines integration for an arbitrary function $f$?

That is, what's wrong with defining an integration of an arbitrary function $f$ as the supremum of $\sum_{x\in s(X)} x\mu(s^{-1}(x))$ over simple meadurable functions $s≦f$?

share|cite|improve this question
up vote 5 down vote accepted

Given a Vitali $E$ set in $[0, 1]$, there are no measurable subsets of $E$ with positive measure (see, e.g. here). Hence if we want to integrate $f = \chi_E$, then we would necessarily have $\int f dm = 0$. But one of the "nice" properties of Lebesgue integration is that any nonnegative function with zero integral must be zero almost everywhere - but this implies $E$ has measure $0$, contradicting that it isn't measurable.

So all of a sudden, a lot of results that rely on the idea of equality almost everywhere, or a property holding up to sets of measure $0$ would break down.

Furthermore, the idea that integration and measure are related would break down fundamentally, were this the case. We'd really like to have that

$$\int \chi_A d\mu = \mu(A)$$

for measures; in fact, it's a general result that the function $A \mapsto \int \chi_A f d\mu$ is a measure for measurable functions $f$ (or taking $f$ identically $1$, giving back the measure $\mu$). But if every function is integrable, every set is measurable; but having regularity of the measure requires that some sets aren't measurable.

share|cite|improve this answer
This 100% resolves my question. Thank you very much:) – user156562 Jun 18 '14 at 4:08
You're very welcome. It's a good question to ask. – user61527 Jun 18 '14 at 4:08
"... but having regularity of the measure requires that some sets aren't measurable" or AC fails. $\hspace{.77 in}$ – Ricky Demer Jun 18 '14 at 8:35

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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