# Lebesgue Integral of Non-Measurable Function

In what follows I'm only considering positive real valued functions.

Everywhere I look about the definition of the Lebesgue integral it is required to consider a measurable function. Why do we not define the integral for non-measurable functions? From what I see we require measurablility of the simple functions that approximate f, not f itself. The definition I'm considering is given a measure space $X$ with measure $\mu$ and a measurable function $f$ we define

$$\int_E f \, \mathrm{d}\mu = \sup_{s \in S} \int_X s \,\mathrm{d}\mu$$ where $S = \{ s : X \to [0, \infty) \mid 0 \le s \le f, s \text{ is simple, measurable} \}$.

For example consider $\mathbb{R}$ with the sigma algebra $\varnothing, \mathbb{R}$ with measure $\mu$ given by $\mu(\varnothing) = 0, \mu(\mathbb{R}) = 1$ and consider $f = \chi_{[0,1]}$ then why can't we say that $$\int_{\mathbb{R}} f \,\mathrm{d} \mu = 0$$ (since the only measurable simple function such that $0\le s \le f$ is $s = 0$) which would follow the definition above? Is this not well defined? In general I'm struggling to see why measurable functions (other than measurable simple functions) are used.

• What you specify is called the "lower integral". There is a corresponding "upper integral" as well. And (guess what) the integrable functions are those for which the upper and lower integrals agree. Dec 26, 2014 at 0:06
• @GEdgar hrm, from my references only this is defined (although I follow what you mean from the Darboux integral), nonetheless "integrable" is defined for measurable $f$ such that $\int f <\infty$ - I'm guessing what you just said (upper a lower integrals agree) means measurable, is that what you meant? Dec 26, 2014 at 0:10
• Also: does the downvoter mind explaining why the downvote? Was this a silly question? Dec 26, 2014 at 0:11
• If upper and lower integrals agree, and are finite, then $f$ is measurable, and in fact integrable. On the other hand, if $f$ is measurable, then your lower integral is the integral, and (if it is finite) $f$ is integrable. Dec 26, 2014 at 0:14
• @GEdgar do you have a reference to this? Also, is finiteness really necessary in your first statement (otherwise the upper and lower integrals agreeing would be equivalent to the function being measurable)? Dec 26, 2014 at 0:18

In addition to the previous advice, note that the function you gave does not "approximate" $f$. An approximating sequence $\{s_n\}$ for $f$ (which $f$ would have iff $f$ were measurable provided the measure space for the domain of $f$ is complete) would need to be within a distance of $\epsilon$ from $f$ for any given $\epsilon >0$. However, the function $s$ you gave as an "approximation" cannot approximate $f$, in the sense that the sequence $\{s_n\}$ with $s_n=s$ $\forall n \in \mathbb{N}$ gets no closer than $1$ from $\chi_{[0,1]}$.

Again, it is important to note that a function $f$ on a complete measure space $X$ is measurable if and only if $f$ is the pointwise limit of some sequence of simple functions -or, trivially, is a simple function itself. Consequently, any sequence of "simple" functions approximating a nonmeasurable function must contain a "simple" function with the characteristic function of a nonmeasurable set as part of its construction. In a sense, this is why you must include the assumption that $f$ is measurable in your definition for the Lebesgue integral.

To wit, recall that the integral of a characteristic function is the measure of the pullback set in your domain; in the case of your $f$, since $f$ is characteristic, the integral, were it to be defined, is the measure of the pullback, $$\int f d\mu = \mu\{f^{-1}\{1\}\}=\mu\{[0,1]\},$$

but you have not defined the measure for $[0,1]$, which is not even in your $\sigma$-algebra; nor can we infer the measure of $[0,1]$ from the definition you gave for your measure space, as your collection of measurable subsets is already a closed $\sigma$-algebra that is $\sigma$-finite under $\mu$ (hence, your measure space cannot even be extended, in the usual Caratheodory way, to include $[0,1]$ with an accompanying well-defined measurement).

As a curiosity tangential to your question, it is possible for nonmeasurable functions to arise from limits of simple functions in a complete measure space, but such a collection of simple functions must be uncountable. For instance, with Lebesgue measure on $\mathbb{R}$, take $f=\chi_V$ to be the characteristic function of the (uncountable and nonmeasurable) Vitali set $V$ on $[0,1]$, and consider the (uncountable) collection of measurable functions $\{\chi_v\}, v\in V$. Then $\chi_V=\sup \{\chi_v\}_{v\in V}$. Were we to define an integral as you wish, then in this case, you may want to say that the integral $\int \chi_V = \sup \{ \int \chi_v \} = 0$. But, again, since $\chi_V$ is itself characteristic, we should then have $\mu(V)=0$, but $\mu(V)$ is not defined for $V$ under the Lebesgue measure.

To address your request for a resource, see Royden's Real Analysis, 4th ed., chapters 17 and 18 (particularly pp. 362-363 were helpful as a reference to me for this post).

• Although I'm not familiar with the pullback (I really only have basic measure theory knowledge) this was very helpful. The first half of the post is what I realized from the initial post but I want to thank you for further showing the issues that are presented. Thanks again! Dec 26, 2014 at 18:35
Also, maybe it is useful to show the bad properties of this lower integral when applied to non-measurable funtions. Take your $\{\varnothing, \mathbb R\}$ example. Let $f = \chi_{[0,1]}$ and $g = 1-f$. Then $\int f = \int g = 0$ but $\int(f+g) = 1$. So even simple linearity fails.
• $f$ and $g$ are both non-measurable according to $\{\varnothing, \mathbb R\}$. Nov 12, 2019 at 20:07