Image of a set of zero measure has zero measure I am studying for my final and got stuck on the following problem from the previous year. I put my attempt below.
Suppose that $I\subset \mathbb{R}$ is an open interval, $f:I\rightarrow \mathbb{R}$ is differentiable on $I$ and its derivative is continuous on $I$. If $a,b\in I$ and $E\subseteq [a,b]$ is of Lebesgue measure zero, show that $f(E)$ is a set of Lebesgue measure zero.
I think that since $E$ is a set of measure zero for every $\varepsilon>0$, there is a countable covering of $E$ with disjoint open intervals $E\subset \cup_{i\geq 1} U_i$ where $|U_i|=d_i$, so that $\sum_{i=1}m^\ast(U_i)=\sum_{i=1} d_i<\varepsilon$, then: 
$$
m^\ast(f(E))\leq \sum_{i=1} m^\ast(f(U_i))\leq \sum_{i=1}|(f(u_{i1}),f(u_{i2}))|=\sum_{i=1}|f(u_{i1})-f(u_{i2})|\leq \sup_{a\leq t\leq b}|f'(t)| \varepsilon,
$$
so $m^\ast(f(E))$ has measure zero as $f'$ is bounded on $[a,b]$.
Howerver, I am not sure why $\sum_{i=1} m^\ast(f(U_i))\leq \sum_{i=1}|(f(u_{i1}),f(u_{i2}))|$.
I am also trying to see why this would imply that for a set $E\subseteq I$ with measure zero, $f(E)$ is a set of measure zero. 
 A: Essentially you are taking $ E\subset U $ with $m^*(U)<\epsilon $ where WLOG $U = \cup_i(a_i,b_i) $ for such disjoint intervals and then estimating 
$$ \sum_i |f(b_i)-f(a_i)| = \sum_i \left|\int^{b_i}_{a_i}f'(t)dt\right| \leq \sum_i \int^{b_i}_{a_i}|f'(t)|dt = \int_U |f'(t)|dt \leq \epsilon \sup |f'(t)| $$
But as you rightly doubted, although $U = \cup_i(a_i,b_i) $, it doesn't necessarily imply $f(U) \subset \cup_i(f(a_i),f(b_i)) $ from which you may conclude from above $$ m^*(f(U))\leq \sum_i |f(b_i)-f(a_i)| \leq \epsilon \sup|f'(t)| $$
However the implication is true for monotonically increasing functions. And infact for your function can be expressed as a difference of two increasing functions, so doing the estimate for increasing functions is good enough.To see this last statement you need to define the total variation, as follows 
$$ g(x) = V^x_a(f) = \sup\{\ \sum_i|f(x_i)-f(x_{i-1})| \ a= x_0 <x_1<...<x_n = x\} $$ Where the sup is taken over all partitions. So you can observe that for $ a < x<y\leq b $ you have 
$$ V^y_a(f) = V^x_a(f) + V^y_x(f) $$ which makes $g $ increasing, and also observe $ |f(y) -f(x)| \leq V^y_x(f) = V^y_a(f)-V^x_a(f) $ and hence $ g(y)-f(y) \geq g(x)-f(x) $ making $ g-f $ also increasing, and $ f = g - (g-f) $ and you can estimate $ g(U) $ and $ (g-f)(U) $ seperately.
