When I studied Measure Theory I learnt it in an axiomatic way, lacking context for why the notion of $\sigma$-algebras was introduced and how they arose in the first place. In this post I want to dispell the magic shrouding their definition by providing motivation for why they are necessary for the most natural conception of measures and the Lebesgue integral.

If you were coming up with your own way for describing the “measure of size” $\lambda(X) \geq 0$ of subsets $X \subset \mathbb{R}$ on top of which a theory for integration can be constructed, in a way to generalize the natural size $b − a$ of an interval $[a, b]$, the following three assumptions/requirements would appear to be entirely natural and expected:

• For any interval $[a, b]$, we have $\lambda([a, b]) = b − a$, and $\lambda(\phi) = 0$; moreover $\lambda(X) \leq \lambda(Y)$ if $X \subset Y$.
• For any subset $X \subset \mathbb{R}$ and $t \in \mathbb{R}$, we have $$\lambda(X_t) = \lambda(X), \text{ where } X_t = t + X = \{ x \in \mathbb{R} | x − t \in X\}$$ (i.e., invariance under translation)
• For any sequence $(X_n)_{n \geq 1}$ of subsets of $\mathbb{R}$, such that $X_n \cap X_m = \phi$ if $n\neq m$, we have $$\lambda\left(\bigcup_{n \geq 1} X_n \right) = \sum_{n \geq 1} \lambda(X_n)$$ where the sum on the right-hand side makes sense in $[0, +\infty]$.

I will now show that it is a consequence of the Axiom of Choice that these three seemingly benign assumptions are impossible to satisfy concurrently, and that is precisely why we are forced to come up with $\sigma$-algebras.

Proof. Consider the quotient $X = \mathbb{R}/\mathbb{Q}$ (i.e., the quotient modulo the equivalence relation given by $x \sim y$ if $x − y \in \mathbb{Q}$ (how do you visualize this? Don't––it's a mess.).). Then, by the Axiom of Choice, there exists a map $$f : X \to \mathbb{R}$$ which associates a single element in it to each equivalence class. By shifting along integers we can, without loss of generality, assume that $\mathfrak{N} = f(X) \subset [0, 1]$ (we replace $f(x)$ by its fractional part if needed). We now look at $\lambda(\mathfrak{N})$ through the lens of our three properties to conclude that our conception is not tenable.
Indeed, by the definition of the equivalence relations, we have $$\mathbb{R} = \bigcup_{t \in \mathbb{Q}} (t + \mathfrak{N}),$$
over the countable set $\mathbb{Q}$. Invariance under translation requires that $\lambda(\mathfrak{N}) = \lambda(t + \mathfrak{N}) \forall t \in \mathbb{Q}$. Thus, if $\lambda(\mathfrak{N}) = 0$ countable additivity would imply that $\lambda(\mathbb{R}) = 0$, which is absurd. Thus, we must have $\lambda(t + \mathfrak{N}) = \lambda(\mathfrak{N}) = c > 0$ for all $t \in \mathbb{Q}$. Now consider the disjoint union, $$\mathfrak{M} = \bigcup_{t \in [0, 1] \cap \mathbb{Q}} (t + \mathfrak{N}).$$
$\mathfrak{M} \subset [0, 2]$ and so $$2 = \lambda([0, 2]) \geq \lambda(\mathfrak{M}) = \sum_{t \in [0, 1] \cap \mathbb{Q}} \lambda(t + \mathfrak{N}) = \sum_{t \in [0, 1] \cap \mathbb{Q}} c = \infty,$$ which is again a contradiction. $\blacksquare$

Not including countable additivity is too drastic an omission to permit a good theory, since it breaks down all limiting arguments. The way around this difficulty has been to restrict the sets for which one tries to define the measure $\lambda(X)$. By considering only suitable “well-behaved sets”, it so turns out that one can avoid the problem above. However, it also turns out that there is no unique notion of “well-behaved sets,” suitable for all sets over which we might want to integrate functions, and therefore one proceeds by axiomatically prescribing the common properties that characterize the collection of well-behaved sets––and thus we arrive at $\sigma$-algebras.