Lebesgue measure of a compact set and a line

113 Views Asked by At

I am trying to work out the problem below. My only idea at the moment is that the compactness of $K$ gives me convergent subsequences. Any help would be appreciated.

Suppose that $K$ is a compact subset of the unit square $[0, 1]^2$ in $R^2$ and the two-dimensional Lebesgue measure of $K$ is greater than 1/2. Show that there are infinitely many lines $L$ in $R^2$ such that $m_L(K ∩ L) > 1/2$, where $m_L$ is the one-dimensional Lebesgue measure on the line $L$.

1

There are 1 best solutions below

0
On BEST ANSWER

Denote by $\mu$ the Lebesgue measure over $[0,1]$. The 2-dimensional Lebesgue measure over $[0,1]^2$ is given by the product measure $\lambda=\mu\times \mu$ and using Fubini we have:

$$\lambda(K)=\int 1_K(x,y) \ d\lambda(x,y)=\int_{0}^1\left (\int_0^1 1_K(x,y)\ d\mu(x)\right ) d\mu(y)$$

Now if there are only finitely many lines $L$ such that $L\cap K$ has length bigger than $\frac{1}{2}$ we would have $$\int_0^1 1_K(x,y)\ d\mu(x)\leq \frac{1}{2}$$

from where $$\lambda(K)\leq\int_{0}^1\frac{1}{2} d\mu=\frac{1}{2}$$ and this contradicts the hypothesis.