Lebesgue Measure and Measure 
My approach: Let F(x) = $μ((0,x])$ for $x >0$
Since from the theorem $F(x) = x F(1)$ for some $x >0$
$μ((0,x])$ $=$ $x$ $μ((0,x])$
 A: Prove that the measures $\mu((0,1])\lambda,\mu$ are equal to the algebra of sets $\mathcal{A}=$ $\{$ finite  unions of intervals $[a,b)$ such that $-\infty \leq a<b \leq +\infty$ $\}$
.
Then since this measures are sigma finite,the by a known theorem they are equal to $\sigma(\mathcal{A})=\mathcal{B}_{\Bbb{R}}$ 
Then use the fact that every Lebesgue measurable set is a union of a borel set and a set of Lebesgue measure zero.
Both measures are equal to sets of measure zero.
A: Let's look at $\mathbb{R}$, the idea for $\mathbb{R}^k$ is the same, albeit messier.
Consider the diadic interval $[0,2^{-n})$ for a fixed $n\in\mathbb{N}$. Note that
$$
[0,1)=\bigcup_{i=0}^{2^n-1} [i2^{-n},(i+1)2^{-n})
$$
Note that each interval $[i2^{-n},(i+1)2^{-n})$ is a translation of $[0,2^{-n})$. Also, these intervals are pairwise disjoint, so
$$
\alpha:=\mu ([0,1))=\sum_{i=0}^{2^n-1} \mu([i2^{-n},(i+1)2^{-n}))=2^n\mu([0,2^{-n}))
$$
Thus
$$
\mu([0,2^{-n}))=\alpha2^{-n}=\alpha\lambda([0,2^{-n}))
$$
So we have the result for the dyadic interval $[0,2^{-n})$. By translation, we also have it for any dyadic interval of length $2^{-n}$. Since $n\in\mathbb{N}$ was arbitrary. the desired conclusion holds for any dyadic interval (of any length). Now, for any open set $U$, $U$ is a disjoint union of some countable collection of dyadic intervals $\langle I_n\rangle_{n=1}^\infty$ so that
$$
\mu(U)=\sum_{n=1}^\infty \mu(I_n)=\sum_{n=1}^\infty \alpha \lambda (I_n)=\alpha \lambda (U)
$$
You'd now need to extend this to closed, $F_\sigma$ and $G_\delta$ sets to complete the argument.
