The rest of the proof runs as before.
Although these proofs run along similar lines, there are subtle adjustments necessary to fit the argument to each new situation.
Let $A'$ be $A$ run backwards.
As $t$ runs from 0 to 1, the point $f(t)$ runs through the interval $[a,b]$.
The problem one runs into, however, is that $f$ need not be smooth.
But this obvious attack runs into a serious difficulty.
But if we argue as in (5), we run into the integral ......, which is meaningless as it stands.
......where $E$ runs over $\langle$runs through$\rangle$ the family $B$.
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