## Measurablity; Inner and Outer area

Let S be the subset of x-axis consisting of the union of the open interval of length 1/4 centered at 1/2, and the open intervals of length 1/16 centered at 1/4 and 3/4, the open intervals of length 1/64 centered at 1/8, 3/8, 5/8, and 7/8, and so forth. Let U = S x (0,1) be the union of the open rectangles of height 1, based on these intervals. Thus U is the union of one rectangle of area 1/4, two rectangles of area 1/16, four rectangles of area 1/64, ... some of which overlap.

(1) Show that U is dense in R and hence the outer area is 1.

It says to use the result that if S is a bounded set in R^2, then S and its closure have the same outer area.

My guess to this problem is to show that U closure is R, and since U is bounded then the outer area is equal to 1.

(2) Let V = R \ U. Show that V is a closed set whose inner area is 0 and whos outer area is bigger than 1/2.

This one I am confused about. It seems clear that R \ U must be a closed set since it includes R's boundary and since U is dense in R, then the inner area must be 0. But how to show outer area is bigger than 1/2?