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Let $B=\left [\frac{l}{2^j},\frac{l+1}{2^j}\right)$ for $j\in\mathbb{N}$ and $0\leq l<2^j$ and $A=\bigcup_{r=0}^{2^n-1}\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right)$ for $k,n\in\mathbb{N}$ with $i\in\mathbb{N}$ fixed, $0\leq k<2^i$.

I tried to show that $A\cap B\neq\emptyset$ if $n>j$ but I was very confused about this problem.

On the other hand, if $m$ is the Lebesgue measure and if $n>j$, is it true that

$m(A\cap B)=m(A)m(B)$?

I tried...

Since the intervals in $A$ are disjoint \begin{eqnarray*} m(A\cap B)&=&m\left( \bigcup_{r=0}^{2^n-1}\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right) \cap \left [\frac{l}{2^j},\frac{l+1}{2^j}\right) \right)\\ &=&\sum_{r=0}^{2^n-1} m\left(\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right) \cap \left [\frac{l}{2^j},\frac{l+1}{2^j}\right) \right) \end{eqnarray*}

but I'm not really sure that

$$m\left(\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right) \cap \left [\frac{l}{2^j},\frac{l+1}{2^j}\right) \right)=\frac{1}{2^{i+n}}\frac{1}{2^j}\;.$$

I tried induction but it did not work. Can somebody help me or give me a hint?

Sorry for my bad english

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    To clarify, is $i$ in $A$ a fixed natural number?2011-12-24
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    yes, $i\in\mathbb{N}$ and fixed, sorry I fotgot it2011-12-24
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    Also, in one set you are iterating over $j$ and in the other $j$ is fixed. Any chance you can clean this up?2011-12-24
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    sorry, was my mistake, in the set $A$ $j$ is a index but I have benn corrected, I exchanged by $r$. I hope the question is more understandable. Other mistake: $0\leq k<2^i$ I had written $0\leq l<2^i$ Sorry for my bad english.2011-12-24

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