Why is the following true? 
Can someone please explain to me the last part? Why does it follow? I am having a hard time seeing why.
Thanks.
PS: $| \cdot | $ is Lebesgue measure. $ A \subseteq \mathbb{R}$ is measurable set.
 A: The fraction you ask about (with $h+k$ in the denominator) is a weighted average of the corresponding fractions in the preceding formulas (i) and (ii), the weights being $h/(h+k)$ and $k/(h+k)$.  So all you need is that, if two numbers are in an interval (in this case the interval $(1-\epsilon,1]$), then any weighted average of those two numbers, being between the two numbers, is also in that interval.
A: The proof is using the fact that if $B$ and $D$ are positive, and if $A/B < z$ and $C/D < z$, then
$\frac{A+C}{B+D} < z$
That on its own isn't very hard to show:
Since $\frac AB < z$ and $B>0$, $A < Bz$.
Since $\frac CD < z$, and $D>0$, $C < Dz$.
Then $A + C < Bz + Dz = (B+D)z$, and $\frac{A+C}{B+D} < z$ (note $B+D>0$).
More generally, $(A+C)/(B+D)$ is always between $A/B$ and $C/D$ on the real number line, or all three expressions could be equal.
A: since the intersection of $[x-k,x] \cap A$ and $[x,x+h] \cap A$ has measure zero, you have:
$$
\mid [x-k,x+h]\cap A \mid = \mid [x-k,x]\cap A \mid + \mid[x,x+h]\cap A \mid
$$
you have also:
$$
k \ge \mid[x-k,x]\cap A\mid \gt k(1-\epsilon) \\
h \ge \mid [x, x+h]\cap A \mid \gt h(1-\epsilon)
$$
adding the inequalities:
$$
k+h \ge \mid[x-k,x]\cap A\mid + \mid [x, x+h]\cap A \mid \gt (k+h)(1-\epsilon)
$$
i.e.
$$
k+h \ge \mid[x-k,x+h]\cap A\mid \gt (k+h)(1-\epsilon)
$$
dividing through by the positive quantity $h+k$ gives the required conclusion
