# 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. – GEdgar Dec 26 '14 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? – DanZimm Dec 26 '14 at 0:10
• Also: does the downvoter mind explaining why the downvote? Was this a silly question? – DanZimm Dec 26 '14 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. – GEdgar Dec 26 '14 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)? – DanZimm Dec 26 '14 at 0:18

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.

• Aha this makes sense - totally missed this. In other words we could define the integral of non-measurable functions but then we couldn't approximate by measurable simple functions and so a lot of niceness of the integral goes away. Thanks! – DanZimm Dec 26 '14 at 0:23
• which function is not measurable here? – David Warren Katz Nov 12 '19 at 18:50
• $f$ and $g$ are both non-measurable according to $\{\varnothing, \mathbb R\}$. – GEdgar Nov 12 '19 at 20:07

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! – DanZimm Dec 26 '14 at 18:35
• This is an excellent answer. I'm surprised it hasn't received more upvotes. – E.O. Nov 1 '16 at 19:50