Nonatomicity and continuity Given a function $f: [0,1) \to \mathbb{R}$ with $f(0)=0$, define a measure $m_f$ on $[0,1)$ by:
$$m_f([0,x)) = f(x)$$
Are the following relations true?


*

*$f$ is a non-decreasing function iff $m_f$ is a non-negative measure;

*$f$ is a continuous function iff $m_f$ is a non-atomic measure;

*$f$ is a differentiable function iff $m_f$ is absolutely continuous with respect to length?


I think I can prove the forward implications:


*

*Suppose $f$ is non-decreasing. So for every half-open interval:


$$m_f([a,b)) = m_f([0,b)) - m_f([0,a)) = f(b) - f(a) \geq 0$$
so $m_f$ is nonnegative.


*Suppose $f$ is continuous. For every $x\in[0,1]$:


$$\{x\} = \bigcap_{n>0} [0,x+{1\over n}) ~~~ \setminus ~~~ [0,x)$$
Therefore:
$$m_f(\{x\}) = \lim_{y\to x+} f(y) - f(x)$$
If $f$ is continuous, then this difference equals 0, so $m_f(\{x\})=0$.


*Suppose $f$ is differentiable. Let $f'$ be its derivative. Then:


$$m_f([0,x)) = f(x) = \int_{y=0}^x f'(y)dy$$
therefore, the measure of any set with length 0 is 0, so $m_f$ is absolutely continuous w.r.t. length. 
Are these proofs correct? Are the opposite directions true too?
 A: First of all, a measure have to be defined on a sufficiently reach family of set. As a most natural choice I put that $m_f$ is a $\sigma$-additive measure defined on a $\sigma$-algebra $\mathcal B$ of all Bored subsets of $[0,1)$ (not including the right endpoint) such that $m_f([0,x))=f(x)$ for each $x\in [0,1]$. In particlar, holds the following 
Lemma. Let $\{A_n\}$ be a sequence of Borel subsets of $[0,1)$. Then $$m_f\left(\bigcap_{n=1}^\infty A_n\right)= \lim_{n\to\infty}m_f\left(\bigcap_{k=1}^{n} A_k\right).$$
Proof. 
$$m_f\left(\bigcap_{n=1}^\infty A_n\right)=m_f\left(A_1\right)-
m_f\left(A_1\setminus\bigcap_{n=1}^\infty A_n\right)
=$$
$$m_f\left(A_1\right)- m_f\left(\bigcup_{n=1}^\infty\left(\bigcap_{k=1}^{n} A_k\setminus\bigcap_{k=1}^{n+1} A_k\right)\right)=\mbox{ (by $\sigma$-additivity)}$$
$$m_f\left(A_1\right)- \sum_{n=1}^\infty m_f\left(\bigcap_{k=1}^{n} A_k\setminus\bigcap_{k=1}^{n+1} A_k\right)=$$
$$m_f\left(A_1\right)- \sum_{n=1}^\infty\left( m_f\left(\bigcap_{k=1}^{n} A_k\right)-m_f\left(\bigcap_{k=1}^{n+1} A_k\right)\right)=$$
$$m_f\left(A_1\right)-\lim_{n\to\infty}\left(m_f\left(A_1\right)- m_f\left(\bigcap_{k=1}^{n+1} A_k\right)\right)=$$
$$\lim_{n\to\infty}m_f\left(\bigcap_{k=1}^{n+1} A_k\right).$$
But $\sigma$-additivity of $m_f$ (which you have exploited in the proof of 2) may fail. For instance, let $f(x)=0$, if $x$ is rational and $f(x)=1$, if $x$ is irrational. If the measure $m_f$ is $\sigma$-additive then by Lemma 
$$m_f(\{0\})=m_f\left(\bigcap_{n=1}^\infty \left[0,\frac 1n\right)\right)= \lim_{n\to\infty} 
m_f\left(\left[0,\frac 1n\right)\right)=\lim_{n\to\infty} f\left(\frac 1n\right)-f(0)=0.$$
On the other hand, 
$$m_f(\{0\})=m_f\left(\bigcap_{n=1}^\infty \left[0,\frac 1{\sqrt{2}n}\right)\right)= \lim_{n\to\infty} 
m_f\left(\left[0,\frac 1{\sqrt{2}n} \right)\right)=\lim_{n\to\infty} f\left(\frac1{\sqrt{2}n}\right)-f(0)=1,$$
a contradiction.
So in order to have $\sigma$-additivity of $m_f$, we have to impose some additional conditions on the function $f$. In particular, in order to avoid the above contradiction, for each $x\in [0,1)$ there have to exist $f(x^+)=\lim_{y\to x+} f(y)$ (*).   
1) This case was considered in The Phenotype’s and fourierwho’s  comments, so we skip it.
2) Assuming $\sigma$-additivity of $m_f$, we can adapt your proof as follows. Pick any $n_0>1/(1-x)$. Then 
$$m_f(\{x\})=m_f\left(\left[x,x+\frac 1{n_0}\right)\setminus\bigcup_{n>n_0} \left[x+\frac 1n,x+\frac 1{n-1}\right)\right)=$$ $$m_f\left(\left[x,x+\frac 1{n_0}\right)\right)-\sum_{n>n_0}m_f\left(\left[x+\frac 1n,x+\frac 1{n-1}\right)\right)=$$ $$f\left(x+\frac 1{n_0}\right)-f\left(x\right)-\sum_{n>n_0}f\left(x+\frac 1{n-1}\right)-f\left(x+\frac 1{n}\right)=$$
$$f\left(x+\frac 1{n_0}\right)-f\left(x\right)-\lim_{n\to\infty}f\left(x+\frac 1{n_0}\right)-f\left(x+\frac 1{n}\right)=(by\mbox{ } *)$$
$$f\left(x+\frac 1{n_0}\right)-f\left(x\right)-f\left(x+\frac 1{n_0}\right)+f\left(x^+\right)= f(x^+)-f(x),$$
that is for each $x\in [0,1)$ we have $m_f(\{x\})=0$ iff $ \lim_{y\to x+} f(y)=f(x)$ (**).   
3) If the function $f$ is absolutely continuous and $m_f$ is a measure then (using Lemma) it is easy to check that $m_f$ is absolutely continuous with respect to the length. There exist absolutely continuous functions which are not differentiable. For instance, a function $f$ such that $f(x)=x$, if $0\le x\le\frac 12$ and $f(x)=1-x$, if $\frac 12\le x\le 1$ is not differentiable at $x=\frac 12$. 
On the other hand, I recall that if $f$ is a differentiable function defined on a segment $[a,b]$ and its derivative $f’$ is Riemann integrable on $[a,b]$ then $f(b)-f(a)= \int_a^b f’(t)dt$ [Fich, 310]. Since a Riemann integrable on $[a,b]$ function is bounded, in this case the function $f$ is absolutely continuous. On the other hand, Lebesgue(?) proved that an absolutely continuous function $f$ defined on a segment $[a,b]$ has a derivative $f’$ almost everywhere, the derivative is Lebesgue integrable, and for each $a\le x\le b$, $f(x)-f(a)= \int_a^x f’(t)dt $. A refined description of the situation is provided in $[BZ]$. “A  fundamental  result  in  real  variable  theory  is  that if  $d$  an  absolutely  continuous  function  on  a  compact  interval  $[a,b]$,  then  $f’$  exists  almost everywhere  and  $|f’ |$  is  integrable  on  $[a,b]$.  The  converse  is  false.  However,  if  it assumed  that  $f’$  exists  everywhere  on  $[a,b]$  and  that  $|f'|$  is  integrable,  then  $f$  is absolutely  continuous.  For  a  particularly  simple  proof  of  this,  see  [Gof]. Through a  personal  communication,  we  have  learned  that  C.  J.  Neugebauer  has  improved this  result  by  assuming  merely  that  the  approximate  derivative  of  $f$  exists  everywhere  and  that  its  absolute  value  is  integrable”. So, I guess that if we take a differentiable function $f$ on $[0,1]$ such that $|f’ |$  is not integrable than both $f$ and $m_f$ may fail to be absolutely continuous. 
References 
[BZ] Thomas  Bagby, William  P.  Ziemer Pointwise  differentiability and  absolute  continuity,  Trans. AMS, 191 (1974), 129-148.
[Fich] Grigorii Fichtenholz Differential and integral calculus, vol. II, 7-th edition, M.: Nauka, 1970 (in Russian).
[Gof]  C.  Goffman On  functions  with  summable  derivatives,  Amer.  Math.  Monthly  78 (1971), 874-875.
A: Answer to 3). If $m_f$ is absolutely continuous we can only say that f is differentiable almost everywhere. So the right question is whether almost everywhere differentiability of f is equivalent to absolute continuity of f. This is false. If $m_f$ is singular then f is differentiable almost everywhere (with derivative 0!). A well known example of such a function is the Cantor funcion. 
A: The typical set up when you want to define a measure by $m_{f}([a,b)) = f(b) - f(a)$ is you assume $f : [0,1) \to \mathbb{R}$ has bounded variation.  Henceforth, I will just say "f is a BV function."  This is not a technicality, as I will try to convince you at the end of this post.  Note that if $f$ is BV, then $m_{f}$ will have finite total variation (i.e. it will take values in $[-R,R]$ for some $R > 0$).  In my answer, I tried to avoid assuming $m_{f}$ has finite variation, but it seems to be hard to give clean answers unless we assume $f$ is non-decreasing.  Therefore, I will assume $f$ is BV and $m_{f}$ has finite variation and try to convince you this is the right approach.  
As was pointed out in another answer, we need to modify the definition of $m_{f}([a,b))$ slightly in order to ensure we get a measure.  Therefore, set $m_{f}([a,b)) = f(b^{-}) - f(a^{-})$, where $f(x^{-}) = \lim_{c \to x^{-}} f(x)$.  This approach is standard: see, for example, this article on BV functions.  If $f$ has finite total variation, then these limits always exist and this procedure ensures that $m_{f}$ extends to a (finite) signed Borel measure on $[0,1)$.  (One can use the Riesz Representation Theorem and Riemann-Stieltjes integrals to prove this, or, alternatively, the Hahn-Caratheodory Extension Theorem.)  
Your argument that $m_{f}$ is non-negative if and only if $f$ is non-decreasing is the right idea (though there are some details to work out once you add the left-hand limits to your definition of $m_{f}$).  I want to correct something I said in a comment: you don't need any "regularity" to complete the argument.  The family $\mathcal{A} = \{A \in \mathscr{B}_{[0,1)} \, \mid \, m_{f}(A) \geq 0\}$ is a monotone class.  You showed it contains all intervals of the form $[a,b)$.  Therefore, $\mathscr{B}_{[0,1)} \subseteq \mathcal{A}$ by the monotone class theorem. 
Again, you have the right idea regarding question (2); there's just a little extra work dealing with the left-hand limits.  (One doesn't need to use properties of BV functions here, contrary to what I wrote previously.)  
Now to address (3):  $m_{f}$ is an absolutely continuous signed measure (with finite variation) if and only if $f$ is an absolutely continuous function (i.e. $f' \in L^{1}([0,1))$ and $f(x) = \int_{0}^{x} f'(y) \, dy$ almost everywhere).  This is a classical result proved, for example, in Royden-Fitzpatrick.   
A question that I didn't think was addressed above is the following: if $f$ has bounded variation and $f'$ exists at every point in $[0,1)$ (or $(0,1)$), is $m_{f}$ absolutely continuous?  Of course, these are two different questions.  I thought this was worth doing because we know what the answer is when $f'$ is Riemann integrable or if $f$ is absolutely continuous, but somehow "$f'$ exists everywhere and $f$ is BV" doesn't obviously fit into either of those.
If $f$ has bounded variation and is differentiable on $[0,1)$ (with $f'(0)$ interpreted as a one-sided derivative), then $m_{f}$ is absolutely continuous.  First, observe that for each $y < 1$, $f$ has finite total variation on $[0,y]$ and properties of BV functions yield
$$\int_{0}^{y} |f'(x)| \, dx \leq TV_{[0,y]}(f) < \infty.$$
Now write $dm_{f} = f'(x) dx + d\nu_{f}$, where $\nu_{f}$ and Lebesgue measure are mutually singular.  For $\nu_{f}$-almost every point $x \in (0,y)$, we have
$$\limsup_{h \to 0} \frac{m_{f}((x - h,x + h))}{2h} = \infty.$$
Thus, $f$ is not differentiable at $\nu_{f}$-almost every point $x \in (0,y)$.  We're assuming $f$ is differentiable on $(0,1)$ so it follows that $\nu_{f} = C \delta_{0}$ for some $C \in \mathbb{R}$.  
Now if $f$ is not differentable at $0$ , there's nothing we can say.  For example, if we take 
$$f(x) = \left\{\begin{array}{r l}
0, & x = 0\\
1, & x > 0.
\end{array} \right.$$
then $f$ is differentiable on $(0,1)$ and has bounded variation, but $m_{f} = \delta_{0}$.  On the other hand, if $f$ is differentiable at $0$, then it is certainly continuous there so our work on (2) implies $C = 0$ and, thus, $m_{f}$ is absolutely continuous.   
Now, why do I insist that $f$ be a BV function?  First, let's define what this means.  The total variation of $f$ on $[0,y]$ is defined by 
$$TV_{[0,y]}(f) = \sup \left\{ \sum_{j = 1}^{N} |f(x_{j}) - f(x_{j - 1})| \, \mid \, 0 = x_{1} < x_{2} < \dots < x_{N} = y, \, \, N \in \mathbb{N} \right\}.$$
We say that $f$ has bounded variation on $[0,y]$ if $TV_{[0,y]}(f) < \infty$.  We say $f$ has bounded variation on $[0,1)$ if there is a constant $C > 0$ such that $TV_{[0,y]}(f) \leq C$ whenever $y < 1$.  If $f$ is non-decreasing, then $TV_{[0,y]}(f) = f(y) -
 f(0)$ so bounded non-decreasing functions on $[0,1)$ have bounded variation.  For better or worse, what goes wrong if $f$ does not have bounded variation is intimately connected to pathologies arising in the theory of signed measures.  (If $f$ is non-decreasing but not bounded, $m_{f}$ will be an infinite non-negative measure and the answers to (1), (2), and (3) are still true with slight modifications.)
An example of a function on $[0,1]$ that doesn't have finite total variation is 
$$f(x) = \sqrt{x} \cos(\pi x^{-1}).$$
The problem is $|f((n + 1)^{-1}) - f(n^{-1})| = \frac{1}{\sqrt{n}} - \frac{1}{\sqrt{n + 1}} \simeq \frac{1}{n}$ so $TV_{[0,1]}(f) = \infty$ by comparison with the harmonic series.  What would happen if I tried to define a signed measure by $m_{f}([0,x)) = f(x)$?  If I write
$$(0,1] = \bigcup_{n = 1}^{\infty} \left[(n + 1)^{-1},n^{-1}\right],$$
then I should have
\begin{align*}
m_{f}((0,1]) &= \sum_{n = 1}^{\infty} m_{f}([(n + 1)^{-1},n^{-1}]) \\
&= \sum_{n = 1}^{\infty} f(n^{-1}) - f((n + 1)^{-1}) \\
&= \sum_{n = 1}^{\infty} \frac{(-1)^{n}}{\sqrt{n(n + 1)}},
\end{align*}
but this last sum is conditionally convergent.  In particular, I get a contradiction when I write
$$m_{f}((0,1])) = \sum_{n = 1}^{\infty} m_{f}([(j_{n} + 1)^{-1},j_{n}^{-1}])),$$
for some other enumeration $(j_{n})_{n \in \mathbb{N}}$ of $\mathbb{N}$ and observe that, by conditional convergence, the right-hand side can be any number.  Things get even worse if I observe that
\begin{align*}
\sum_{n = 1}^{\infty} m_{f}([(2n + 1)^{-1},(2n)^{-1}]) &= \sum_{n = 1}^{\infty} \frac{1}{2 \sqrt{n (n + \frac{1}{2})}} = \infty
\end{align*}
while
\begin{align*}
\sum_{n = 1}^{\infty} m_{f}([(2n)^{-1},(2n - 1)^{-1}) &= - \sum_{n = 1}^{\infty} \frac{1}{2 \sqrt{n (n - \frac{1}{2})}} = -\infty,
\end{align*}
which also yields $m_{f}((0,1]) = \infty - \infty$.
EDIT:  We can extend our class of functions $f$ slightly if we consider $f : [0,1) \to \mathbb{R}$ satisfying $f = f_{1} - f_{2}$, where $f_{1},f_{2} : [0,1) \to \mathbb{R}$ are non-decreasing and either $f_{1}$ or $f_{2}$ is bounded.  Then $m_{f} = m_{1} - m_{2}$, where $m_{1}$ and $m_{2}$ are the measures obtained from $f_{1}$ and $f_{2}$ respectively.  Notice that $m_{1}$ and $m_{2}$ are non-negative measures and one of them is finite so $m_{f}$ is well-defined.  The answers to questions (1)-(3) follow directly from the corresponding answers for $f_{1}$ and $f_{2}$.  
