In constructing the Lebesgue measure we start by defining the outer measure of a set in terms of coverings of the set by countable collection of cubes.
Suppose we instead start by approximating sets from the inside. Say, by defining the "inside measure" of a set as the supremum of the total volume of any countable collection of disjoint cubes contained in the set.
I can see that we would run into problems. For example on $[0,1]$ the set of rationals and set of irrationals would both have "inside measure" $0$ whereas the interval as a whole would have "inside measure" $1$.
What I would like to have is an intuitive understanding of why we run into these problems when approximating from inside whereas approximating from outside gives us a measure theory which works.
The reason I am asking is that for simple sets, for eg. when calculating the area of a circle, approximating from outside and approximating from inside works equally well.
Best Answer
If you approximate from outside you get an outer measure that is subadditive: $\mu^\ast[A \cup B] \le \mu^\ast[A] + \mu^\ast[B]$. This allows to think of measure as some kind of a "norm" of a set, and, in particular, define a metric $\rho(A,B) := \mu^\ast[A \vartriangle B]$, where $\vartriangle$ is symmetric difference. Completion and extension by continuity using this metric gives you measure theory (by this I mean that, first of all, the measure algebra is just the completion of the space of "simple" sets by this metric, and measurable sets are exactly those that can be approximated by "simple" ones using this metric).
Now if you try to approximate from inside, you get an inequality in the wrong direction: for $A$ and $B$ disjoint $\mu[A \cup B] \ge \mu[A] + \mu[B]$. So there is no metric, no completion, ..., no measure theory. In particular, this doesn't give any straightforward way to select which sets are measurable and map them to elements of the measure algebra.