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?
Why don’t we just use the outer measure
integrationmeasure-theory
Best Answer
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 $[0,1]$.