$g(x) = x^3$, if $N$ is a null set on $\mathbb{R}$, can we conclude that $g^{-1}(N)$ is also a null set? 
$g(x) = x^3$, if $N$ is a null set on $\mathbb{R}$, can we conclude that $g^{-1}(N)$ is also a null set?

I think the answer is yes, and my attempt is as follows:
Let $N_k = N \bigcap [k,k+1]$, then $N$ = $\bigcup_{k \in \mathbb{z}} N_k$. It suffices to show that $\forall k, g^{-1}(N_k)$ is null set. Let $m$ denote the Lebesgue measure. Consider the case in which $k \geq 1$. If $(\alpha,\beta) \subset [k,k+1]$, then 
\begin{gather*}
m(g^{-1}((\alpha,\beta))) &= \beta^{\frac 13} - \alpha^{\frac 13}
  = \frac{\beta - \alpha}{\alpha^{\frac 23} + \beta^{\frac 23} + \alpha^{\frac 13}\beta^{\frac 13}}
  \leq \frac{\beta - \alpha}{3\alpha^{\frac 13} \beta^{\frac 13}}
  \leq \frac{\beta - \alpha}{3} 
\end{gather*}
By open set construction theorem on $\mathbb{R}$, we have that for any open set $G \subset [k,k+1]$, 
$$
  m(g^{-1}(G)) \leq \frac 13 m(G)
$$
Because $N_k$ is null set, so $\forall \epsilon > 0$, $\exists$ open set $G \subset [k,k+1]$, such that $m(G) < 3\epsilon$. Since $g$ is continuous on $\mathbb{R}$, $g^{-1}(G)$ is open and $g^{-1}(N_k) \subset g^{-1}(G)$, and 
$$
  m(g^{-1}(G)) \leq \frac 13 m(G) < \epsilon
$$
Hence $N_k$ is null set.
However, I am struggling with the proof that $g^{-1}(N_0)$, where $N_0 \subset [0,1]$, is a null set. Because in the inequality
\begin{gather*}
m(g^{-1}((\alpha,\beta))) &= \beta^{\frac 13} - \alpha^{\frac 13}
  = \frac{\beta - \alpha}{\alpha^{\frac 23} + \beta^{\frac 23} + \alpha^{\frac 13}\beta^{\frac 13}}
  \leq \frac{\beta - \alpha}{3\alpha^{\frac 13} \beta^{\frac 13}}
\end{gather*}
$\alpha$ and $\beta$ can be very close to 0. So is there alternative method to show the statement is true? Or actually the statement is false? Any hint will be appreciated!
 A: You are almost done; just  realize that your proof for $[k,k+1]$ applies verbatim to every interval $[a,b]$ with $0<a<b$. No need to restrict yourself to integer endpoints.
The halfline $(0,\infty)$ can be written as a countable union of such intervals, say 
$$(0,\infty) = \bigcup_{k=1}^\infty \left[\frac{1}{k},k\right] \tag{1}$$
(I don't even care that these overlap a lot). 
For $N\subset (0,\infty)$, let 
$$N_k = N\cap \left[\frac{1}{k},k\right]$$
By your proof, $g^{-1}(N_k)$ is a null set. Then
$$g^{-1}(N) = \bigcup_{k=1}^\infty g^{-1}(N_k)$$
is a countable union of null sets, hence a null set.
The same works for $(-\infty,0)$. And $g^{-1}(0)=\{0\}$ is just one point.
The idea of (1) is sometimes called the method of exhaustion: a big open set is written as the union of compact sets.
A: For any function, absolutely continuous on $\mathbb{R}$, image of set of measure 0 is a set of measure 0. In your case the absolutely continuous function is $g^{-1}(x)=\sqrt[3]x$. See Luzin N-property: http://en.wikipedia.org/wiki/Luzin_N_property
