Let $\mathcal{L}$ be the set of oriented lines in $\mathbb{R}^2$ and let $\mu$ be the Kinematic Measure on $\mathcal{L}$; up to scaling, $\mu$ comes from the unique (up to scaling) volume form on $\mathcal{L}$ that is invariant under the action of the group of isometries of $\mathbb{R}^2$. Given any function $f:[0,1] \rightarrow \mathbb{R}^2$, define a function $n(f):\mathcal{L} \rightarrow \mathbb{Z} \cup \{\infty\}$ by setting $n(f)(L)$ to be the number of elements of the set $\{\text{$x \in [0,1]$ $|$ $f(x) \in L$}\}$.

The key step in one of the standard proofs of Crofton's formula asserts the following. Let $f:[0,1] \rightarrow \mathbb{R}^2$ be a rectifiable curve. For any partition $P$ of $[0,1]$, let $f_P:[0,1] \rightarrow \mathbb{R}^2$ be the associated polygonal approximation to $f$, so $f_P(x) = f(x)$ for $x \in P$ and $f_P$ restricts to a straight line segment on each component of $[0,1] \setminus P$. The set of partitions of $[0,1]$ is partially ordered by refinement, and by definition the length of $f$ is the limit (with respect to this partial order) of the lengths of the $f_P$.

The "key step" referred to above is the following: the limit (over the set partitions $P$ of $[0,1]$) of $\int_{\mathcal{L}} n(f_P) d\mu$ equals $\int_{\mathcal{L}} n(f) d\mu$.

I don't see how to prove this and cannot find a source that has a complete proof. Can anyone either give me a proof or a good reference for it?