# Calculating explicit outer and inner Lebesgue measures for the half-open interval $E = [1,3)$?

I am trying to get a better understanding of the definitions of Lebesgue inner and outer measures. Let $E$ be an arbitrary set of real numbers, and for any open interval $I=(a,b)$ define $\lambda(I)=b-a$.

• The Lebesgue outer measure $\lambda^\ast$ is defined as $$\lambda^\ast(E) = \inf\{\lambda(G):E \subset G, G \ \text{is open}\}.$$
• The Lebesgue inner measure is defined as $$\lambda_\ast(E) = \sup\{\lambda(F):F \subset E, F \ \text{is compact}\}.$$
• If these measures of $E$ are equal we say that $E$ is Lebesgue measurable with the Lebesgue measure $\lambda(E) = \lambda^\ast(E) = \lambda_\ast(E)$.

Now lets say I have the half open interval $E=[1,3)$. Clearly the measure of this set is $2$. But How can I formally use the definitions above to calculate $\lambda^\ast(E)$ and $\lambda_\ast(E)$ and show that they equal $2$?

I think that there is something missing in your statement, since you define $$\lambda$$ only on open intervals, but you do not say how it is defined on open sets or compact sets. One usually defines the Lebesgue outer measure as
$$\lambda^*(E) = \inf \sum_n |Q_n|,$$
where the infimum is taken over countable covers of $$E$$ with closed cubes, and $$|Q_n|$$ is the elementary geometric volume. A set is said to be Lebesgue measurable if for all $$\varepsilon >0$$ there is an open set containing $$E$$ with $$\lambda^*(O - E) \le \varepsilon$$ and for such sets the Lebesgue measure is defined as the exterior measure.
Okay, then $$E \subset (1-1/n,3)$$ and $$\lambda(1-1/n,3)=2+1/n$$, hence $$\lambda^*(E) \le 2$$. Moreover, $$[1+3-1/n] \subset E$$ and $$\lambda([1+3-1/n])=2-1/n$$ (this is already a closed interval), hence $$\lambda_*(E) \ge 2$$ and you are done since $$\lambda_* \le \lambda^*$$ (the smallest closed interval containing a compact set is contained in the smallest open interval containg that set).