If $f[a,b]\to\mathbb{R}$ is Lipchitz and $X\subset [a,b]$ has measure zero show that $f(X)$ has measure zero too.
What I did:
As $X$ has measure zero, $\forall \epsilon>0$ there exists a countable union of open intervals $I_i$ s.t. $$X\subset \cup_{i=1}^\infty I_i$$ and $\sum_{i=1}^\infty |I_i|<\epsilon.$ As $f$ is continuous $f(I_i)=J_i$ is an interval and $f(X)\subset \cup_{i=1}^\infty J_i$. As $f$ is Lipschitz $|J_i|\leq k|I_i|$, for some $k>0.$ Now $\sum_{i=1}^\infty |J_i|<k\sum_{i=1}^\infty |I_i|<k\epsilon,$ and so one can take $\epsilon=\frac{\epsilon_1}{k}$ to finish.
Right?
Best Answer
Image of open interval doesn't need to be open (for example, $\sin((-10, 10)) = [-1, 1]$). But image of interval of measure less than $x$ anyway can be covered with interval of measure less than $kx$.
If $I_i$ is an open interval with measure less than $x$ then $\overline{I_i}$ is closed interval with measure less than $x$, image $f(\overline{I_i})$ of $\overline{I_i}$ is closed interval with measure less than $kx$, and closed interval with measure less than $y$ can be covered with open interval with measure less than $y$.