# Inner measure doesn't care whether finite or countable

I was reading this answer, to try understanding why the inner measure defined as the $$\mu_*(U)=\sup\{\mu(A):A\subset U\text{ is measurable}\}$$ is not really used. I copied the answer here :

Tao talks about this in his measure theory book, which last time I checked was still freely available.

He says roughly that there is an asymmetry here because when measuring finite unions of boxes, the measure is subadditive and not superadditive. This leads to the fact that the Jordan inner measure $$m^{*, J}(U) := \sup \{m(A)\ :\ A \subset U\ \text{is a finite union of boxes}\}$$ doesn't care whether or not you put finite or countable in the definition (because you are taking a supremum and its subadditive). So a natural Lebesgue inner measure where you put countable here gives no increase in power/resolution of the measure

Does this mean that $$m^{*, J}_1(U) := \sup \{m(A)\ :\ A \subset U\ \text{is a finite union of boxes}\}$$ and $$m^{*, J}_2(U) := \sup \{m(A)\ :\ A \subset U\ \text{is a countable union of boxes}\}$$

are the same? I don't understand how to use the subadditivity of the supremum to show this.

I was thinking about this today while reading the Tao's book and have this proof which I'm not sure whether it's correct or not (also this proof is not related to subadditivity of elementary measure which Tao said is the reason) Here is the proof, and please let me know if this is wrong:

let's assume $$E\subset\mathbb{R}^d$$ , $$C=\{m(A)|A=\bigcup\limits_{i=1}^{\infty}A_{i}; A\subset{E}\}$$ which $$A_i$$ are boxes in $$\mathbb{R}^d$$. $$C$$ is the set of all measures using countable unions of boxes. Also, $$F=\{m(A)|A=\bigcup\limits_{i=1}^{n}A_{i}; A\subset{E}\}$$. $$F$$ is the set of all measures using finite unions of boxes. now, assume $$sup(C)>sup(F)$$. if this is true then, there exist a countable union of boxes, that its measure is bigger than every element of set $$F$$; let's call it $$K$$, assume $$m(K)- sup(F) = t \implies m(K)-m(A)\geq t\,, \forall m(A) \in F$$.(*). we know $$K=\bigcup\limits_{i=1}^{\infty}K_{i}$$. consider sets $$f_n=\bigcup\limits_{i=1}^{n}K_{i} \implies m(f_n) \in F$$; we know $$m(f_n)\to m(K)$$ as $$n\to\infty$$ and this is contradiction becasue we have $$\forall \epsilon>0, \, \exists N\in\mathbb{N} , s.t. |m(f_n)-m(K)|<\epsilon; \forall n\ge N$$. now if we choose $$\epsilon\ll t$$ we have contradiction because of *.