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enter image description here

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.

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  • $\begingroup$ downvoter: What is wrong with this question? please, explain. $\endgroup$ – ILoveMath Dec 9 '13 at 0:17
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    $\begingroup$ i don't downvote, but it would be helpful if you stated what the set $A$ is. without that information the question is difficult to interpret $\endgroup$ – David Holden Dec 9 '13 at 0:18
  • $\begingroup$ ok just edited. $\endgroup$ – ILoveMath Dec 9 '13 at 0:19
  • $\begingroup$ This is a duplicate of question you asked a short while ago (except in the other question you omitted pertinent information). math.stackexchange.com/questions/597855/… $\endgroup$ – copper.hat Dec 9 '13 at 2:50
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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

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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.

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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.

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    $\begingroup$ can you show why this result is true? thanks $\endgroup$ – ILoveMath Dec 9 '13 at 0:28
  • $\begingroup$ Note: $A,B,C,D$ are positive here. $\endgroup$ – GEdgar Dec 9 '13 at 0:30
  • $\begingroup$ Added a short proof. $\endgroup$ – aschepler Dec 9 '13 at 1:01

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