Is the Function Phi Lipschitz for Lebesgue Measurable Sets?

[mathmari]

Hey!

Let $E \subset \mathbb{R}^d$ Lebesgue measurable and $\phi (t)=m \left ( \Pi_{i=1}^{d} (-\infty , t_i ) \cap E \right )$. To show that $\phi$ is Lipschitz, can we do it as followed??

Let $x>y$.

$$|\phi(x) - \phi(y)|=|m \left ( \Pi_{i=1}^{d} (-\infty , x_i ) \cap E \right )-m \left ( \Pi_{i=1}^{d} (-\infty , y_i ) \cap E \right )|=|m \left [ \left ( \Pi_{i=1}^{d} (-\infty , x_i ) \cap E \right ) \setminus \left ( (-\infty , y_i ) \cap E \right ) \right ]| \leq m \left ( \Pi_{i=1}^{d} [y_{i}, x_{i} ] \right )=\Pi_{i=1}^{d} [y_{i}, x_{i}]$$

Is it correct?? (Wondering)

 

[Jhoneine]

Yes, that is correct. You can also see that $\phi$ is Lipschitz by noting that the difference $|\phi(x) - \phi(y)| \leq |x-y|$, which follows from the fact that $E$ is measurable and that $m(\cdot)$ is a measure.