Why do we care so much to restrict to a $\sigma$-algebra of sets on which the outer measure is countably additive? When is countable sub-additivity not enough? Does it pose a problem when defining the Lebesgue integral?

  • 2
    $\begingroup$ Countable additivity is perhaps the most important property of a measure. You can show that the outer measure is additive when any two sets are separated by positive distance, but we desire additivity beyond that. $\endgroup$ Oct 13, 2020 at 2:58
  • 1
    $\begingroup$ I'm starting a bounty because I think this is a good question. Specifically, I think the question is, "what goes wrong if you try to define integration against an outer measure rather than a measure?" If nothing goes wrong, then why not formulate measure theory in terms of outer measures and forego all the complicatedness of measurability? $\endgroup$
    – N. Virgo
    Jun 25, 2023 at 0:58
  • $\begingroup$ @N.Virgo See the construction of Lebesgue integral in Royden’s Real Analysis (Proposition 3 on pp.79 3e for me) $\endgroup$
    – Andrew
    Jun 25, 2023 at 1:02

2 Answers 2


Below, $\mu^*$ denotes Lebesgue outer measure and $\mu$ denotes its restriction to the measurable sets.

Remember how measurability enters the definition of the Lebesgue integral of a (for simplicity) nonnegative function $f$ on $[0,1]$: we look at functions $h$ which have $h(x)\le f(x)$ for all $x\in[0,1]$, take on only finitely many values, and have measurable preimages of each particular value. Each such $h$ has a corresponding "naive integral" $$\theta_h:=\sum_{y\in im(h)}y\cdot\mu\{x\in [0,1]: h(x)=y\},$$ and we define $\int_{[0,1]}fd\mu$ to be the supremum of these $\theta_h$s.

If we drop that last condition, we get garbage: for example, let $f(x)=1$ everywhere, let $(V_i)_{i\in\mathbb{N}}$ be a decomposition of $[0,1]$ into Vitali sets of equal outer measure $\epsilon>0$, and for each $n$ let $$g_n(x)={1\over \min\{n, v(x)\}}$$ where $v(x)$ is the unique $i$ such that $x\in V_i$. (Note that I'm assuming $0\not\in\mathbb{N}$ here for simplicity; if, like me usually!, you prefer $0$ to be a natural number, throw in a "$+1$" appropriately.)

On the one hand we always have $g_n(x)\le f(x)$ for all $n$ and $x$. On the other hand the "naive integral" of each $g_n$ is $\ge \sum_{i=1}^n{\epsilon\over n}$, which is a big problem since the harmonic series diverges! Basically, if we were to ignore measurability concerns in the definition of the Lebesgue integral, every positive function would wind up having infinite integral over $[0,1]$.

  • $\begingroup$ This makes sense, though it is interesting to me that the counterexample is highly non-constructive. $\endgroup$ Jun 25, 2023 at 2:40
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    $\begingroup$ @CharlesHudgins Well, it's consistent with $\mathsf{ZF + DC}$ (= set theory with a weak form of the axiom of choice) that every set is measurable, so that's unavoidable. (This is a theorem of Solovay.) $\endgroup$ Jun 25, 2023 at 2:58
  • $\begingroup$ Thank you for the comment. I didn't know a result like that existed. I knew that non-measurable sets were non-constructive, but didn't know they were consistent with a weakened form of choice. $\endgroup$ Jun 25, 2023 at 3:05
  • $\begingroup$ *their nonexistence is consistent with a weakened form of choice. $\endgroup$ Jun 25, 2023 at 3:21
  • $\begingroup$ @Noah Schweber: After careful and multiple reads, I understand a good amount of your answer ( that is pretty shocking in itself ). One question I do have is why you have to define $g_{n}(x)$ as the $\frac{1} {(min(n, \nu(x)})$ rather than just $\frac{ 1} {\nu(x)}$. Thanks. $\endgroup$
    – mark leeds
    Jul 3, 2023 at 1:28

Instead of viewing the domain of the outer measure as being restricted from $P(\mathbb{R})$ down to $\mathcal{B}$ in order to achieve countable additivity we could view it from the opposite direction, by observing that the outer measure is naturally countably additive already on quite a large class of sets - namely the `system of intervals' $\mathcal{I} = \{$ all countable unions of intervals in $\mathbb{R}\ \}$ - a proof of this is given from the definition of outer measure in [1]. However we find that the class of sets $\mathcal{I}$ lacks a crucial property required for the domain of the outer measure - namely closure under set complementation (a counterexample is provided in the Appendix of [1]), which then motivates the idea of a $\sigma$-algebra and the Borel algebra $\mathcal{B}$. We extend the domain $\mathcal{I}$ to $\mathcal{B}$ to achieve the closure under set complementation and then find that the countable additivity property is preserved (eg see [2, Chap 2]).

However a further reason for requiring countable additivity is that it allows certain useful theorems on outer measure to be proved - as Axler [2, §2C, p41] says :

The countable additivity that forms the key part of the definition [of a measure] above allows us to prove good limit theorems.

To see how these 'good limit theorems' come about let us first go back to some basic intuitive ideas about what we would expect from a notion of measure on $\mathbb{R}$ :

  1. We require some function $\mu : \mathcal{C} \rightarrow [0, \infty]$ defined on some class of sets $\mathcal{C}$ in $\mathbb{R}$ which contains all intervals, and for which $\mu(I) = l(I)\ \forall$ intervals $I$, ie $\mu$ extends the notion of length of an interval in $\mathbb{R}$ to some larger class of sets.
  2. We require $\mu(D \cup E) = \mu(D) + \mu(E)$ when $D, E$ are disjoint and more generally finite additivity ie. $\mu(E_1 \cup \cdots \cup E_n) = \mu(E_1) + \cdots + \mu(E_n)$ when $E_1, \ldots, E_n$ are disjoint.
  3. We require order preservation : ie. $D, E \in \mathcal{I}$ and $D \subseteq E \Rightarrow \mu(D) \leq \mu(E)$ (follows from 2)
  4. $\mu(\emptyset) = 0$ (follows from 2)

Countable additivity and closure of $\mathcal{C}$ under set complementation are not properties we would initially expect of a measure. However as we see below both these properties are needed to prove certain desired theorems involving limits of measures.

Firstly, from the above properties we have the following limiting processes :

If $E_1 \subseteq E_2 \subseteq \cdots$ is an increasing sequence of sets in $\mathcal{C}$ then : \begin{equation} \lim_{n \rightarrow \infty} \mu(\bigcup_{k=1}^{n} E_k ) = \lim_{n \rightarrow \infty} \mu(E_n) = \sup \{ \mu(E_n) \} \in [0, \infty] \label{eq:limit-union} \tag{1} \end{equation}

and, if $E_1 \supseteq E_2 \supseteq \cdots$ is a decreasing sequence of sets in $\mathcal{C}$ then : \begin{equation} \lim_{n \rightarrow \infty} \mu(\bigcap_{k=1}^{n} E_k ) = \lim_{n \rightarrow \infty} \mu(E_n) = \inf \{ \mu(E_n) \} \in [0, \infty] \label{eq:limit-intersection} \tag{2} \end{equation}

It is then natural to ask the question whether in (\ref{eq:limit-union}) and (\ref{eq:limit-intersection}) it could make sense for the limit operation to be 'passed through' the function $\mu$ to give the following :

If $E_1 \subseteq E_2 \subseteq \cdots$ is an increasing sequence of sets in $\mathcal{C}$ then : \begin{equation} \bigcup_{n=1}^{\infty} E_n \in \mathcal{C}\ \hspace{1em} \mbox{and} \hspace{1em} \mu(\bigcup_{n=1}^{\infty} E_n) = \lim_{n \rightarrow \infty} \mu(E_n) \label{eq:limit-union-pass-through} \tag{3} \end{equation}

and, if $E_1 \supseteq E_2 \supseteq \cdots$ is a decreasing sequence of sets in $\mathcal{C}$ then : \begin{equation} \bigcap_{n=1}^{\infty} E_n \in \mathcal{C}\ \hspace{1em} \mbox{and} \hspace{1em} \mu(\bigcap_{n=1}^{\infty} E_n) = \lim_{n \rightarrow \infty} \mu(E_n) \label{eq:limit-intersection-pass-through} \tag{4} \end{equation}

Note that since any union $\bigcup_{n=1}^{\infty} E_n$ can be expressed as an increasing union $\bigcup_{n=1}^{\infty} (E_1 \cup \cdots \cup E_n)$, (\ref{eq:limit-union-pass-through}) requires that $\mathcal{C}$ must be closed under all countable unions (and hence contain all countable unions of intervals). And since any intersection $\bigcap_{n=1}^{\infty} E_n$ can be expressed as a decreasing intersection $\bigcap_{n=1}^{\infty} (E_1 \cap \cdots \cap E_n)$, (\ref{eq:limit-intersection-pass-through}) requires that $\mathcal{C}$ must be closed under all countable intersections (and hence contain all countable intersections of intervals).

However in addition to the above heuristic to motivate the limit properties (\ref{eq:limit-union-pass-through}) and (\ref{eq:limit-intersection-pass-through}), a concrete example where properties (\ref{eq:limit-union-pass-through}) & (\ref{eq:limit-intersection-pass-through}) solve a problem in real analysis outside of measure theory can be found in the Bounded Convergence Theorem for Riemann Integrals in Bartle [3, §22.14, p288], where the author has omitted a portion of the proof which can be supplied by properties (\ref{eq:limit-union-pass-through}) & (\ref{eq:limit-intersection-pass-through}). In Bartle & Sherbert [4, §7.2.5, p304], this BCT is also stated but the proof is omitted entirely, being described as 'quite delicate'. What is required is to show that if $\delta > 0$ and if $(E_n)$ is a sequence of sets in $[0, 1]$ with each $E_n$ containing a finite number of non-overlapping closed intervals with a total length $\geq \delta$, then there exists a point belonging to infinitely many of the $E_n$. This can be proved using (\ref{eq:limit-union-pass-through}) & (\ref{eq:limit-intersection-pass-through}) (together with the finite additivity, order preservation, and extension of interval length) as follows : we require to show that the set $E = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty} E_m$ is non-empty, so letting $F_n$ denote the union of the above intervals within $E_n$ (so $F_n \in \mathcal{C}$ due to (\ref{eq:limit-union-pass-through})), and letting $F = \bigcap_{n=1}^{\infty} \bigcup_{m=n}^{\infty} F_m \subseteq E$, then we have from (\ref{eq:limit-intersection-pass-through}), $\mu(F) = \lim_{n \rightarrow \infty} \mu(\bigcup_{m=n}^{\infty} F_m)$, where $\forall\ n,\ \mu(\bigcup_{m=n}^{\infty} F_m) \geq \mu(F_n) \geq \delta$ (noting $F_n$ equals a finite union of disjoint intervals with a total length $\geq \delta$), so that $\mu(F) > 0$, and hence $F \neq \emptyset$ and thus $E \neq \emptyset$. In Bartle [3] the Bounded Convergence Theorem for Riemann integrals is subsequently used to prove the Dominated Convergence Theorem for Riemann integrals [3, §25.21, p359], which in turn is used to prove Stirling's Formula in [5]. Thus we see a number of important concrete applications of properties (\ref{eq:limit-union-pass-through}) & (\ref{eq:limit-intersection-pass-through}).

Consider now how we can prove the properties (\ref{eq:limit-union-pass-through}) and (\ref{eq:limit-intersection-pass-through}). Firstly note (\ref{eq:limit-union-pass-through}) must hold if any $\mu(E_n) = \infty$, by order preservation, so for (\ref{eq:limit-union-pass-through}) assume every $\mu(E_n) < \infty$. Since the $E_n$ are increasing, the sets $E_n \setminus E_{n-1}$ are disjoint. Thus provided each $E_n \setminus E_{n-1} \in \mathcal{C}$ and $\mathcal{C}$ is closed under countable union we would obtain : \begin{eqnarray} \mu(\bigcup_{n=1}^{\infty} E_n) & = & \mu(\bigcup_{n=1}^{\infty} (E_n \setminus E_{n-1}) ), \hspace{2em} \mbox{defining } E_0 = \emptyset \nonumber \\ & = & \sum_{n=1}^{\infty} \mu(E_n \setminus E_{n-1}), \hspace{2em} \mbox{if countable additivity applied} \nonumber \\ & = & \lim_{n \rightarrow \infty} \sum_{k=1}^{n} \mu(E_k \setminus E_{k-1}) \label{eq:telescope} \tag{5} \end{eqnarray}

But again applying countable additivity we would have within $\overline{\mathbb{R}}$ : $$\mu(E_k) = \mu(E_{k-1}) + \mu(E_k \setminus E_{k-1})$$ and so as every $\mu(E_k)$ is in $\mathbb{R}$ : $$\mu(E_k \setminus E_{k-1}) = \mu(E_k) - \mu(E_{k-1})$$

so that (\ref{eq:telescope}) telescopes to give the required (\ref{eq:limit-union-pass-through}). The countable additivity was essential at two critical points of the argument. We note the identity $\mu(E \setminus D) = \mu(E) - \mu(D)$ holds whenever $\mu(D) < \infty$.

In the above, as well as the countable additivity of $\mu$ and the closure of $\mathcal{C}$ under countable union we required that the set difference $E_n \setminus E_{n-1} \in \mathcal{C}$ for any $E_n$ and $E_{n-1}$ in $\mathcal{C}$. Since $\mathcal{C}$ contains all intervals, and hence $\mathbb{R}$ itself, we thus require in particular that $\mathbb{R} \setminus E \in \mathcal{C}, \forall\ E \in \mathcal{C}$, ie. $\mathcal{C}$ is closed under set complementation. From De Morgan's laws the latter together with closure under countable union implies closure under countable intersection.

Property (\ref{eq:limit-intersection-pass-through}) readily follows on from (\ref{eq:limit-union-pass-through}), with no additional properties of the measure being required, but the condition $\mu(E_1) < \infty$ is required :

Since $(E_n)$ is decreasing $(E_1 \setminus E_n)$ is increasing and so applying (\ref{eq:limit-union-pass-through}) we have : \begin{eqnarray*} \mu(\bigcup_{n=1}^{\infty} E_1 \setminus E_n) & = & \lim_{n \rightarrow \infty} \mu(E_1 \setminus E_n) \\ \mbox{ie.} \hspace{3em} \mu(E_1 \setminus \bigcap_{n=1}^{\infty} E_n) & = & \lim_{n \rightarrow \infty} \mu(E_1 \setminus E_n), \hspace{2em} \mbox{by De Morgan.} \end{eqnarray*} But $$\mu(E_1 \setminus \bigcap_{n=1}^{\infty} E_n) = \mu(E_1) - \mu(\bigcap_{n=1}^{\infty} E_n)$$ and $$\mu(E_1 \setminus E_n) = \mu(E_1) - \mu(E_n)$$ so $$\mu(E_1) - \mu(\bigcap_{n=1}^{\infty} E_n) = \lim_{n \rightarrow \infty} (\mu(E_1) - \mu(E_n)) = \mu(E_1) - \lim_{n \rightarrow \infty} \mu(E_n)$$ therefore, since $\mu(E_1) < \infty$, (\ref{eq:limit-intersection-pass-through}) follows.

Thus in summary, without the countable additivity property we would not be able to obtain the required limit theorems (\ref{eq:limit-union-pass-through}) and (\ref{eq:limit-intersection-pass-through}), and we can see the necessity of this property prior to considering the Lebesgue integral. In summary the properties required of $\mu : \mathcal{C}\ \rightarrow [0, \infty]$ for it to satisfy both the above intuitive notions of a measure and the desired limit theorems are :

(A) For the domain $\mathcal{C}$ of $\mu$ :

  1. contains all intervals
  2. is closed under countable union
  3. is closed under set complementation

(B) For the function $\mu$ :

  1. satisfies $\mu(I) = l(I)\ \forall$ intervals $I \subseteq \mathbb{R}$
  2. is countably additive, ie $\mu( \bigcup_{n=1}^{\infty} E_n ) = \sum_{n=1}^{\infty} \mu(E_n)$, whenever $(E_n)$ are disjoint sets in $\mathcal{C}$

In Axler [2] proof that countable subadditivity follows on from these properties is given in Theorem 2.58, and the above limit theorems are given as Theorems 2.59 and 2.60.


[1] Ross Ure Anderson (2023), Intervals and Outer Measure on $\mathbb{R}$, https://ia801904.us.archive.org/19/items/intervals-and-outer-measure-on-r/IntervalsAndOuterMeasureOnR.pdf

[2] Sheldon Axler (2020), Measure, Integration & Real Analysis, Springer Graduate Texts in Mathematics, https://measure.axler.net/.

[3] Robert G. Bartle (1964), The Elements of Real Analysis, John Wiley & Sons.

[4] Robert G. Bartle & Donald R. Sherbert (1982), Introduction to Real Analysis, John Wiley & Sons.

[5] Keith Conrad, Stirling's Formula, https://kconrad.math.uconn.edu/blurbs/analysis/stirling.pdf


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