# Measure my balls

Suppose $X$ is a compact metric space and $\mu$ is a finite Borel measure on $X$. Is the measure $\mu$ determined by the value of the measure of the balls?

# Measure this

Let $\mu,\nu$ be non-negative, finite Borel measures on $\mathbb{R}^n$ that are singular with respect to each other. Find $\displaystyle\lim_{r\to 0}\dfrac{\nu(B(x,r))}{\mu(B(x,r))}$ for $\mu$-almost every $x$ and $\nu$-almost every $x$.

# Too large to measure

Let $X$ be an infinite dimensional normed space. There is no non-trivial translation invariant Borel measure on $X$ which is finite on open balls.

# Slicing up compact sets

Show that if $K$ is a compact subset of $\Bbb R^n$, and if every intersection of $K$ with an hyperplane $x_n=a$ has measure zero in $\Bbb R^{n-1}$, then $K$ has measure zero.

Note This is valid for any measurable set in any $\sigma$-compact measure space by virtue of Tonelli’s theorem. The point is to prove this without using this. The fact that $K$ is compact will come in handy.