# Proving that a subset of a Jordan measurable set is also Jordan measurable.

Suppose that you have $S \subseteq T \subseteq \mathbb{R^n}$, and that $T$ is Jordan measurable with $m(T) = 0$ as well.

Prove that $S$ is also Jordan measurable with $m(S) = 0$.

I have attempted to prove this, and I am unsure if I am correct. Hence I could use feedback on my proof.

S is jordan measurable with measure 0 if for every $\epsilon > 0$ there exists a finite collection of covers {$U_1$, $U_2$, ..., $U_n$} such that $\sum_{i=1}^n V(U_i) < \epsilon$.

Since every $x_i \in S$ and $S \subseteq T$, then $x_i \in T$. Since T is measure 0, there exists a collection of covers {$O_1$, ..., $O_n$}, so that $T \subseteq \bigcup_{i=1}^n O_i$.

Since $S \subseteq T$, then $S \subseteq\bigcup_{i=1}^n O_i$

Take $U_i$ = $O_i$, so that $O_i$ covers an open neighbourhood around $X_i$ in S.

Hence,

$\sum_{i=1}^n V(U_i) = \sum_{i=1}^n V(O_i) < \epsilon$

More over, since S is measure 0, it is Jordan measurable.

You seem to turn a bit in circles at the end: Your open cover $O_i$ is already a finite open cover for any subset $S$ of $T$, you don't need to rename it to $U_i$.
Also, it is not clear what you mean by "covers an open neighbourhood around $X_i$". At that point you are basically done and your definition of "Jordan measurable with measure zero" applies immediately, no need for more explanations.