Show that $\mu$ is unique and right-continuous This problem is a little tricky, so I'd like some of your thoughts on it!
Assume $\mu$ is a measure on $(\mathbb{R},\mathbb{B})$ where $\mathbb{B}$ are the Borel sets on $\mathbb{R}$. Also, $\mu((a,b]) < +\infty$ whenever $-\infty < a \leq b < \infty$. Show that there exists a function $F: \mathbb{R} \to \mathbb{R}$ such that $$\mu((a,b]) = F(b) - F(a)$$ and that $F$ is uniquely determined up to a constant. Show that $F$ is right-continuous.
By the fundamental theorem of calculus we know that if $f$ is a real-valued continuous function on $[a,b]$ and $F$ is an antiderivative of $f$ in $[a,b]$, then $$\int_a^b f(t) \, \mathrm{d} t = F(b) - F(a).$$ This function is determined up to a constant, as $\int f(t) \, \mathrm{d}t = F(x) + C$, $C \in \mathbb{R}$. But we know that $\mu$ must be a non-negative set function. How do I fix the integral so that it still satisfied the fundamental theorem of calculus and becomes non-negative? My idea was $\int_a^b |f(t)| \, \mathrm{d}t$, making $f \in L^1(\mathbb{R})$. The integral clearly satisfies that it is equal to $0$ on the empty set and is $\sigma$-additive, so all I need is non-negativity.
To prove right-continuity, I must show that for every $\varepsilon > 0$ there exists a $\delta > 0$ such that for all $h$, $0 < h < \delta$, we have $$|F(x_0+h) - F(x_0)|\leq \varepsilon.$$ But doesn't this follow from the way we chose the function $F$ by integraion? As $\int_{x_0}^{x_0+h} f(t) \, \mathrm{d}t \to 0$ as $h \to 0$.
Hope you can help!
 A: For right continuity we need to show that $\lim_{x \rightarrow x_0^+} F(x) - F(x_0) = \lim_{x \rightarrow x_0^+} \mu(x_0, x] = 0$, where the first equality follows by definition of $F$.
Consider $A_n = (x_0, x_0 +\frac{1}{n}]$, and $A = \cap_{n \rightarrow \infty} A_n = (x_0, x_0] = \emptyset$. We know that $\mu(A) = \mu(\cap_{n=1}^{\infty} A_n)$ since $\mu(A_1) < \infty$ and $A_1 \supset A_2 \supset ...$ We have that $\lim_{n \rightarrow \infty} \mu(A \setminus A_n) = 0$, or equivalently $\lim_{n\rightarrow \infty} \mu(A_n) = \mu(A)$. However, this tells us precisely that $\lim_{x \rightarrow x_0^+} F(x) - F(x_0) = \mu(x_0,x_0] = 0$, as we desired. 
Note, it is not necessarily continuous, because if we approached $x_0$ from the left, then $\cap_{n=1}^{\infty}( x_0 - \frac{1}{n} , x_0] = \{x_0\}$ and there could be finitely many points where the measure $\mu$ acted like a $\delta$-measure and assigned a non-zero measure to a single point.
As for non-negativity of your integral, $f$ is chosen so that $\int_{a}^{b} f(t) dt = F(b) - F(a) = \mu(a,b] \ge 0.$ Therefore, the non-negativity of the integral is already guaranteed by the way $f$ is chosen. However, not every measure on $\mathbb{R}$ can be represented as a Riemann Integral. There are 3 parts to a general measure in $\mathbb{R}$. There is the usual part that can be represented as a Riemann-integral, there is the Cantor Dust, and the $\delta$-measures. Delta measures can be represented by the Riemann-Stieltjes integral, however I don't think in general even this can represent Cantor Dust. Therefore, I present a different solution.
Assume $F$ is not uniquely determined up to a constant. Then, there exists $F, G$ such that $\forall a \le b \in \mathbb{R}$ Such that $F(b) - F(a) = \mu(a,b] = G(b) - G(a)$ and $F(x) - G(x) \neq C, \forall C \in \mathbb{R}$. Then by this last assumption, there exists $x,y$ so that $G(x) - F(x) \neq G(y) - F(y)$. If without loss of generality, $x \le y$, then this implies that $\mu(x,y] = G(y) - G(x) \neq F(y) - F(x) = \mu(x,y]$, a contradiction.
