Let $(X, \mathcal{M}, \mu)$ be a measure space. Suppose $E \in \mathcal{M}$ and $f \in L^+$ where $L^+$ is a space of measurable functions from $X$ to $[0, \infty]$. $\int_E f$ is defined by $\int_X f\chi_E$ where $\chi_E$ is a characteristic function of $E$. Now every time we want to use some property that is true for integrals over whole space $X$, we have to do manipulations with $\chi_E$. For example, let $f_n$ be a sequence in $L^+$ such that $f_n(x) \nearrow f(x)$ for all $x \in E$. Suppose we know that monotone convergence theorem is true for integrals over $X$ and we want to prove $$\int_E f = \lim_{n\to\infty} \int_E f_n.$$ Since $f_n\chi_E \nearrow f\chi_E$ we have $$\int_E f = \int_X f\chi_E = \lim_{n\to\infty} \int_X f_n\chi_E = \lim_{n\to\infty} \int_E f_n.$$ I know it's easy but is there a way to avoid such manipulations? I want to know that equality above and many other statements about integrals are true because $E$ is a measure space in its own right (seems much more natural).

More precisely, for $E \in \mathcal{M}$ define $$\mathcal{M}_E = \{ E \cap F \mid F \in \mathcal{M} \}.$$ It's easy to check that $\mathcal{M}_E$ is a $\sigma$-algebra. $(E, \mathcal{M}_E, \mu|_{\mathcal{M}_E})$ is a measure space and $\int_E f$ in space $(X, \mathcal{M}, \mu)$ is equal to $\int_E f|_E$ in space $(E, \mathcal{M}_E, \mu|_{\mathcal{M}_E})$ for every measurable $f$.

Is this a commonly accepted way of handling integrals over subset of measure space? If not, do I really have to use $\chi_E$ every time or there is some other way?

  • $\begingroup$ You are just using the measure $\mu|_E$, which leads to the same things. As long as you know this measure carries on the properties you want, you're fine. $\endgroup$ – Silvia Ghinassi Jul 1 '16 at 14:33
  • $\begingroup$ @SilviaGhinassi intuitively it's very simple, but could you be more precise? $\endgroup$ – edubrovskiy Jul 1 '16 at 14:56
  • $\begingroup$ I agree with you, and that's why I am afraid of getting into technicalities. I might easily say something wrong, so I'll leave it to someone else to provide you a good answer. $\endgroup$ – Silvia Ghinassi Jul 1 '16 at 15:01

There are basically two interesting types of functions

$$f:X \to [0,\infty] , \mu-\text{ measurable} $$


$$f : X \to [- \infty, \infty] \text{ which is integrable, i.e. } \int f(x) \mu(dx) < \infty .$$

These functions are interesting, because you can define for both the integral $\int_X f(x) \mu(dx)$, but maybe the integral is infinity in the first case.

Many theorems are true for either non-negative measurable functions (e.g. monotonic convergence theorem) or integrable functions (dominant convergence theorem).

Now if you have proven the monotonic convergece theorem for a general measurable space, then you don't need to prove it for integrals restricted on a set, simply because $(E, \mathcal{M}_E, \mu|_{\mathcal{M}_E})$ is again a measure space, and thus the monotonic convergence theorem is also true here. As well as all other properties and theorems that are known for integrals on a measurable space.

So in your case,if you have non-negative functions $f_n(x) \nearrow f(x) $ for all $x\in E$ you immediately get

$$\int f(x) \mu|_{\mathcal{M}_E}(dx) = \lim_{n\to \infty} \int f_n(x) \mu|_{\mathcal{M}_E}(dx) $$

However, to be sure that the theorem really holds for all functions restricted to $E$ you need to understand the relation between integrable functions of $\mu_{\mathcal{M}_E}$ and functions $f$ restricted from $X$ to $E$. The relation is as follows:

Let $f:X \to [0,\infty]$ be $\mu$-measurable, or let $f:X \to [-\infty,\infty]$ be $\mu$ integrable. Define for some set $E \in \mathcal{M}$ the function $f':A \to [-\infty, \infty]$ with $f'(x) := f(x)$. Then, we have

$$ \int f'(x) \mu_{\mathcal{M}_E}(dx) = \int_A f(x) \mu(dx) $$

  • $\begingroup$ Could you give me a link to some book or paper where this approach is used? I've already checked relation between integrals on $X$ and on $E$. I just wanted to know that it's a commonly accepted approach. $\endgroup$ – edubrovskiy Jul 2 '16 at 7:06
  • $\begingroup$ @edubrovskiy I can recommend the book **Measure and Integration Theory ** from Heinz Bauer. Its an old, solid and my personal favorite book. The relation between integrals on $X$ and integrals on $E$ is discussed at the end of §12. $\endgroup$ – Adam Jul 2 '16 at 8:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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