Show that it is Lipschitz Let $E \subset \mathbb{R}^d$ be Lebesgue measurable und let $\phi (t)=m \left ( \Pi_{i=1}^{d} (-\infty , t_i ) \cap E \right )$. 
I have to show that $\phi $ is Lipschitz. 
Could you give me some hint how I could do that??
Let $x>y$.
$$|\phi(x) - \phi(y)|=|m \left ( \Pi_{i=1}^{d} (-\infty , x_i ) \cap E \right )-m \left ( \Pi_{i=1}^{d} (-\infty , y_i ) \cap E \right )|$$
 A: If $E$ has infinite measure, then $\phi$ can have some nasty jump, like for $E = \{ x\in \mathbb{R}^d : x_1 > 0\}$, when $\phi(t) = 0$ if $t_1\leqslant 0$ and $\phi(t) = +\infty$ if $t_1 > 0$.
If $E$ has finite measure, then $\phi$ is continuous, but it need not be even locally Lipschitz if $d > 1$ (for $d = 1$, it is then clearly Lipschitz continuous with Lipschitz constant $1$, since for $h \geqslant 0$ we have $\lvert\phi(t+h)-\phi(t)\rvert \leqslant m(E\cap [t,t+h]) \leqslant h$). As an example, consider
$$E = \left\{(x,y)\in\mathbb{R}^2 : y < 0, 0 < x < \frac{1}{1+y^2}\right\}.$$
Then for $0 < t < 1$ we have
$$\phi(t,0) = t\sqrt{\frac{1}{t}-1} + \arctan \frac{1}{\sqrt{\frac{1}{t}-1}} = \sqrt{t(1-t)} + \arctan \sqrt{\frac{t}{1-t}}$$
which behaves like $2\sqrt{t}$ for small positive $t$, hence is not locally Lipschitz in any neighbourhood of $(0,0)$.
If $E$ is bounded, say $E\subset [-K,K]^d$, then $\phi$ is Lipschitz continuous: Keeping all coordinates except the $i$-th fixed, we have
\begin{align}
0 &\leqslant \phi(t+h\cdot e_i) - \phi(t)\\
&= m(E \cap \mathbb{R}^{i-1}\times [t_i,t_i+h]\times \mathbb{R}^{d-i})\\
&\leqslant m([-K,K]^d \cap  \mathbb{R}^{i-1}\times [t_i,t_i+h]\times \mathbb{R}^{d-i})\\
&\leqslant m([-K,K]^{i-1}\times [t_i,t_i+h] \times [-K,K]^{d-i})\\
&= (2K)^{d-1}\cdot h
\end{align}
for $h \geqslant 0$. Thus $\lvert \phi(t) - \phi(s)\rvert \leqslant (2K)^{d-1}\lvert t-s\rvert$ if $s$ and $t$ differ in at most one coordinate. From this you can get the Lipschitz continuity of $\phi$ by iterating over the coordinates where the two arguments differ.
