
  Let $A,B$ be measurable subsets of $[0,1]$ where $m(A)=m(B)=1/4$.
  Prove that there exists a $t\in\mathbb{R}$ such that $m(A\cap B_{t})>1/1000$.
  $B_{t}$ denotes the translation of $B$ by $t$. 


Perhaps some kind of set approximation works? as measurable sets can be approximated by "simple" sets (finite unions of intervals)

There are several things you can do. Here are four:

You can set up a matrix and use Gaussian elimination to figure out the dimension of the space they span. They span $\mathbb{R}^3$ if and only if the rank of the matrix is $3$. For example, you have
$$\begin{align*}
\left(\begin{array}{ccc}
1 & 1 & 1\\
3 & 2 & 1\\
1 & 1 & 0\\
1 & 0 & 0
\end{array}\right) &\rightarrow \left(\begin{array}{ccc}
1 & 0 & 0\\
3 & 2 & 1\\
1 & 1 & 0\\
1 & 1 & 1
\end{array}\right) &&\rightarrow \left(\begin{array}{ccc}
1 & 0 & 0\\
0 & 2 & 1\\
0 & 1 & 0\\
0 & 1 & 1
\end{array}\right)\\
&\rightarrow \left(\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 2 & 1\\
0 & 1 & 1
\end{array}\right) &&\rightarrow \left(\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 1
\end{array}\right).
\end{align*}$$
(Sequence of operations: exchanged rows 1 and 4; subtracted first row from other rows to make $0$s in first column; exchanged second and third rows; added multiples of the second row to third and fourth row to make $0$s in the second column).

At this point, it is clear the rank of the matrix is $3$, so the vectors span a subspace of dimension $3$, hence they span $\mathbb{R}^3$.

See if one of your vectors is a linear combination of the others. If so, you can drop it from the set and still get the same span; then you'll have three vectors and you can use the methods you found on the web. For example, you might notice that $(3,2,1) = (1,1,1)+(1,1,0)+(1,0,0)$; that means that
$$\mathrm{span}\Bigl\{(1,1,1),\ (3,2,1),\ (1,1,0),\ (1,0,0)\Bigr\} = \mathrm{span}\Bigl\{(1,1,1),\ (1,1,0),\ (1,0,0)\Bigr\}.$$

Determine if the vectors $(1,0,0)$, $(0,1,0)$, and $(0,0,1)$ lie in the span (or any other set of three vectors that you already know span). In this case this is easy: $(1,0,0)$ is in your set; $(0,1,0) = (1,1,0)-(1,0,0)$, so $(0,1,0)$ is in the span; and $(0,0,1) = (1,1,1)-(1,1,0)$, so $(0,0,1)$ is also in the span. Since the span contains the standard basis for $\mathbb{R}^3$, it contains all of $\mathbb{R}^3$ (and hence is equal to $\mathbb{R}^3$).

Solve the system of equations
$$\alpha\left(\begin{array}{c}1\\1\\1\end{array}\right) + \beta\left(\begin{array}{c}3\\2\\1\end{array}\right) + \gamma\left(\begin{array}{c}1\\1\\0\end{array}\right) + \delta\left(\begin{array}{c}1\\0\\0\end{array}\right) = \left(\begin{array}{c}a\\b\\c\end{array}\right)$$
for arbitrary $a$, $b$, and $c$. If there is always a solution, then the vectors span $\mathbb{R}^3$; if there is a choice of $a,b,c$ for which the system is inconsistent, then the vectors do not span $\mathbb{R}^3$. You can use the same set of elementary row operations I used in 1, with the augmented matrix leaving the last column indicated as expressions of $a$, $b$, and $c$.


