# Lebesgue measure #1

Thu 29 May 2014

Filed under math

Tags math

Here is a fun proof that illuminates an important property of Lebesgue measure. Note that the function $m()$ refers to measure.

### Let $X$ be a closed subset of $[0,1]$ such that $m(X) = 1$. Prove that $X = [0, 1]$.

Let $X$ be closed subset of $[0,1]$ such that $m(X) = 1$. We need to show that $X= [0,1]$. Let $a=inf(X)$ and let $b=sup(X)$. Suppose, by contradiction, that $X \neq [0, 1]$. Then, there are 3 cases for which $X=[0, 1]$.

Case 1: $a > 0$. By definition of measure, $m(X) = (b -a) - m([a, b]\backslash X) = 1$, but $(b - a) < 1$ and $m(X)$ cannot equal $1$.

Case 2: $b > 1$. Again, be definition of measure, we fnd that $(b - a) < 1$ and $m(X)$ cannot equal $1$.

Case 3: $a = 0$, and b = $0$, since $b - a = 1$, and since $X = [0, 1]$, then $[0, 1]$ \ $X = 0$ which implies that $X$ must be equal to $[0, 1]$.