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Steinhaus theorem

Let $\lambda$ be Borel measure.
  1. If $A$ is measurable and $\lambda(A)>0$, then $A-A=\left\{ x-y\; :\; x,y\in A \right\}$ contains a segment $I$ such that $0\in I$
  2. If $A, B$ are measurable sets and $\lambda(A), \lambda(B)>0$, then $A+B$ contains a segment I.

Proof

Part 1

Let $f=1_{A}*1_{-A}$, then $f$ is is a convolution of two continuous function, therefore $f$ is continuous. Clearly $f(0)=\int 1_{A}(0-y)1_{-A}(y)dy=\int 1_{A}(-y)a_{-A}(y)dy=\int 1_{A}(-y)\cdot 1_{A}(-y)dy = \int 1_{A}(y)dy=\lambda(A)>0.$ Now since $f$ is continuous in $0$ and $f(0)>0$, we get an interval $I=(-\epsilon,\epsilon)$ around $0$ such that $f(x)>0$ for all $x\in I$.
Let $x_0$ be any member in $\left( A-A \right)^\complement$, then for all real number $t$ we get $\Big( -t\not\in A \mbox{ or } x_{0}-t\not\in A\Big)$, hence $\Big( 1_{A}(x_{0}-t)=0 \mbox{ or } 1_{-A}(t)=0\Big)$. Therefore for any real $t$ we get $1_{A}(x_{0}-t)\cdot 1_{-A}(t)$ and from this we get $f(t)=\int 1_{A}(x_{0}-t)\cdot 1_{-A}(t)dt=0$, so $x_0\in \left\{ x\; : \; f(x)=0 \right\}$ and since $x_0$ was taken as any member in $(A-A)^{\complement}$, we get $(A-A)^{\complement}\subset \left\{ x\; :\; f(x)=0 \right\}$. Now since $\forall x\in I\big( f(x)>0\big)$, we get $I\cap \left( A-A \right)^{\complement}=\emptyset$, so $I\subset A-A$. $\blacksquare$

Part 2

Denote $A=\bigcup\limits_{n=1}^{\infty}A\cap [-n,n]$. Since $\lambda(A)>0$ we get the existence of positive integer $n_{0}>0$ such that $\lambda(A\cap [-n_{0},n_{0}])>0$. Now let $A_{0}= A\cap [-n_{0},n_{0}]$, obviously $\lambda(A_{0})>0$ is finite.
Similarly, one may denote $B=\bigcup\limits_{n=1}^{\infty}B\cap [-n,n]$, since $\lambda(B)>0$ we get the existence of positive integer $m_0$ such that $B_{0}= B\cap [-m_{0},m_{0}]$ and $0<\lambda(B_{0})<\infty$. Now clearly \[ A_{0}+B_{0}\subset A+B.\] Denote \[ f:\mathbb{R}\to 1_{A_{0}}*1_{B_{0}}=\int\limits 1_{A_0}(x-t)1_{B_{0}(t)}dt. \] Since $f$ is continuous (theorem) the set $f^{-1}\Big((0,\infty)\Big)=\left\{ x\; :\; f(x)>0 \right\}$ is open. In addition, we get \[ \int\limits f(x)dx=\lambda(A_{0})\cdot\lambda(B_{0})>0, \] hence, the set $f^{-1}\big((0,\infty)\big)\neq \emptyset$ and open, therefore it must contain nonempty interval $I$. If $x\not\in A_{0}+B_{0}$, then for all real $t$ we get $\Big( t\in B_{0} \mbox{ or } x-t\not\in A_{0}\Big)$ and this implies that $\Big(1_{B_{0}}(t=0) \mbox{ or } 1_{A_{0}}(x-t)=0\Big)$, so $f(x)=0$. Since $x$ is any member in $(A_{0}+B_{0})^{\complement}$, we get $(A_{0}+B_{0})^{\complement}\subset \left\{ x\; :\; f(x)=0 \right\}$ and from this we get $\left\{ x\; : \; f(x)=0 \right\}^{\complement}\subset A_{0}+B_{0}$, so $\left\{ x\; : \; f(x)\not= 0 \right\}\subset A_{0}+B_{0}$ and \[ I\subset f^{-1}\big( (0,\infty)\big)\subset \left\{ x\; : \; f(x)\not= 0 \right\} \subset A_{0}+B_{0} .\] $\blacksquare$