Is it true that $m(A\cap B)=m(A)m(B)$? 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
 A: I will assume that $n>j$ as you seem to be saying this in your question.
I'm afraid that you're approach won't work, as $m(I\cap J)\neq m(I)m(J)$ for the individual intervals $I,J$, with a simple counterexample provided by $I = J = [0,1/2)$. The way I would approach this problem would be to figure out how many intervals $\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right)$ are contained in $\left[\frac{l}{2^j},\frac{l+1}{2^j}\right)$ and add up these measures, then figure out the measures of the at most two intervals of the form $\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right)\cap \left[\frac{l}{2^j},\frac{l+1}{2^j}\right)$ which are nonempty but not the entire first interval. I will denote their measures $m_1,m_2$.
The first of these two intervals corresponds to a value of $r$ such that $$\frac{k+r2^i}{2^{i+n}}\leq \frac{l}{2^j} < \frac{k+1+r2^i}{2^{i+n}}$$ which we successively rearrange as $$k+r2^i\leq \frac{l2^{i+n}}{2^j} < k+1+r2^i$$ 
$$r\leq l2^{n-j} - k2^{-i} < r+2^{-i}$$
which can never be satisfied. To see this, note that $n>j\implies l2^{n-j}\in \mathbb{N}$, so we would need $x - 2^{-i}< k2^{-i}\leq x$ for some $x\in \mathbb{N}$, yet this means $x2^i - 1 < k \leq x2^i$ and so since $k\in \mathbb{N}$, $k = x2^i$, violating $k < 2^i$. Thus $m_1 = 0$ as the intersection of the intervals is empty.
The second of these two intervals corresponds to a value of $r$ such that $$\frac{k+r2^i}{2^{i+n}}< \frac{l+1}{2^j} \leq \frac{k+1+r2^i}{2^{i+n}}$$ which we successively rearrange as $$k+r2^i< \frac{(l+1)2^{i+n}}{2^j} \leq k+1+r2^i$$ 
$$r< (l+1)2^{n-j} - k2^{-i} \leq r+2^{-i}$$
which can never be similarly never be satisfied, so $m_2 = 0$ as well.
Al that remains is to count the number of intervals $\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right)$ contained in $\left[\frac{l}{2^j},\frac{l+1}{2^j}\right)$. Each such interval corresponds to a value of $r$ satisfying $$\frac{l}{2^j}\leq \frac{k+r2^i}{2^{i+n}} < \frac{k+1+r2^i}{2^{i+n}} < \frac{l+1}{2^j}$$ and since we already showed no interval $\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right)$ intersects but is not contained in $\left[\frac{l}{2^j},\frac{l+1}{2^j}\right)$, we can simplify this to
$$\frac{l}{2^j}\leq \frac{k+r2^i}{2^{i+n}} < \frac{l+1}{2^j}$$
$$\frac{l}{2^{j-i-n}}\leq k+r2^i < \frac{l+1}{2^{j-i-n}}$$
$$l{2^{n-j}} - k2^{-i}\leq r < (l+1)2^{n-j} - k2^{-i}$$
and since $0 < k2^{-i} < 1$ we have exactly $2^{n-j}$ possible values for $r$. Furthermore, all of these possible values for $r$ are actually iterated over in $A$, as $n>j,l < 2^j\implies (l+1){2^{n-j}} - k2^{-i} < 2^n$ so $r\leq 2^n - 1$ as at most the greatest integer less than $(l+1){2^{n-j}} - k2^{-i}$. Thus our final answer is
$$m(A\cap B) = 2^{n-j}m\left(\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right)\right) = 2^{n-j}\frac{1}{2^{i+n}} = \frac{1}{2^{i+j}}$$
EDIT: The measure of $A$ is $2^nm\left(\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right)\right) = 2^n\frac{1}{2^{i+n}} = \frac{1}{2^i}$, while the measure of $B$ is clearly $\frac{1}{2^j}$, so we do in fact have $m(A\cap B) = m(A)m(B)$.
A: Fix $i,j,k,l,n\in\mathbb{N}$ with $l<2^j$ and $n>j$. For $0\le r<2^n$ let
$$A_r=\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right)\;,$$
and consider
$$A_r\cap B=\left[\frac{k+r2^i}{2^{i+n}},\frac{k+1+r2^i}{2^{i+n}}\right) \cap \left [\frac{l}{2^j},\frac{l+1}{2^j}\right)\;.$$
Let $m=i+n-j>0$. Then $$B=\left[\frac{l2^m}{2^{i+n}},\frac{l2^m+2^m}{2^{i+n}}\right)\;,$$ so either $l2^m\le k+r2^i<l2^m+2^m$ and $A_r\subseteq B$, or $A_r\cap B=\varnothing$. Thus, $m(A_r\cap B)$ is either $2^{-(i+n)}$ or $0$.
Improved version from this point on:
Let $d=n-j$. Then $l2^m\le k+r2^i<l2^m+2^m$ iff $$l2^d\le \dfrac{k}{2^i}+r<(l+1)2^d\;.\tag{1}$$
Moreover, $0\le k<2^i$, so $0\le \dfrac{k}{2^i}<1$, $\left\lfloor\dfrac{k}{2^i}+r\right\rfloor= r$, and $(1)$ holds iff $$l2^d\le r<(l+1)2^d\;.\tag{2}$$
Clearly $(2)$ holds for precisely $2^d$ values of $r$, all of which are permissible: $l<2^j$, so $l+1\le 2^j$, and $(l+1)2^d\le 2^{d+j}=2^n$. It follows that $m(A\cap B)=2^d2^{-(i+n)}=2^{-(i+j)}$.
Since $m(B)=2^{-j}$ and $m(A)=2^n2^{-(i+n)}=2^{-i}$, it is indeed the case that $m(A\cap B)=m(A)m(B)$.
