Lebesgue measure inequality

Question

I have some trouble in proving that if C is a closed set and A an open set (both in $R^n$ with euclidean metric) with $C \subseteq A$, then $ |C|\le |A|$ where |#| indicates the Lebesgue measure. Edit: obviously A is bounded

This is my attempt: $d(C,R^n ∖ A)\gt 0$ because the function $x→d(x,R^n∖A)$ defined on the compact set C is continuous, therefore it has a minimum m(>0). Let H be a pluri-rectangle that contains C, if we divide it in n-rectangles each one with diameter less than m/2 and we sweep aside those that do not have elements of C we obtain a pluri-rectangle H' contained in A and that contains C. Hence by definition of measure of open and closed sets we have $|C|\le|H'|\le|A| $. Is it correct? thanks in advance

Answer 1

Let $B:=A \setminus C$. Prove that

B is open, hence measurable, and $A=C \cup B$ and that $C \cap B= \emptyset$.

It follows that $|A|=|C|+|B| \ge |C|$.