Showing that a Borel Measure $\mu\equiv 0$ 
Problem. Let $\mu$ be a Borel measure on $[0,1]$. Assume that
  
  
*
  
*$\mu$ and Lebesgue measure $m$ are mutually singular.
  
*$\mu([0,t])$ depends continuously on $t$.
  
*$f\in L^{1}(\mu)$ for any function $f:[0,1]\rightarrow\mathbb{R}$, with $f\in L^{1}(m)$. (Note that $f$ has a finite value at every point.)
  
  
  Show that $\mu\equiv 0$.

This question was asked before, but it did not receive a full answer--just a commented suggestion on how to proceed. 
My issue is that I don't see why $\mu=\delta_{0}$ (i.e. the Dirac mass at $x=0$) is not a counterexample to the problem statement. $\delta_{0}$ and Lebesgue measure are mutually singular. $\mu([0,t])\equiv 1$, which is trivially continuous. And for any function $f$ satisfying 3. $\int f\mathrm{d}\mu=f(0)$, which is finite by hypothesis. Am I missing something obvious?
I believe I can show that under the hypotheses of the problem, we must have $\mu(\left\{0\right\})=\mu([0,1])$. Indeed, suppose that $\mu(\left\{0\right\})<\mu([0,1])$. Since $\mu$ and $m$ are mutually singular, there exist disjoint measure sets $A,B$ with $A\cup B=[0,1]$, $m(A)=0$, and $\mu(B)=0$. Since $F(t):=\mu([0,t])$ is continuous, the intermediate value theorem (IVT) applies. By repeated application of IVT, we can inductively find an increasing sequence $\left\{t_{n}\right\}$ with 
$$t_{0}:=0, \quad \alpha_{n+1}:=\mu([0,t_{n+1}])>\alpha_{n}:=\mu([0,t_{n}])$$
Note that it follows from the continuity of $F$ that $\mu(\left\{t\right\})=0$ for any $0<t\leq 1$. In particular, $\mu([t_{n},t_{n+1}))=\alpha_{n+1}-\alpha_{n}$.
Now define a Borel measurable function $f:[0,1]\rightarrow\mathbb{R}$ by
$$f(s):=\sum_{n=0}^{\infty}\dfrac{1}{\alpha_{n+1}-\alpha_{n}}\chi_{A\cap [t_{n},t_{n+1})}(s)$$
$f\in L^{1}(m)$, since $m(A)=0$, so $f\in L^{1}(\mu)$ by problem assumption. But then for any positive integer $N$,
$$\int f(s)\mathrm{d}\mu(s)\geq\sum_{n=0}^{N}\dfrac{1}{\alpha_{n+1}-\alpha_{n}}\mu(A\cap [t_{n},t_{n+1}))=\sum_{n=0}^{N}\dfrac{1}{\alpha_{n+1}-\alpha_{n}}\mu([t_{n},t_{n+1}))=\sum_{n=0}^{N}1=N$$
Whence, we arrive at a contradiction.
 A: Not an answer, a comment too long for the comment box:
Oh that's you - diidn't notice the name. Didja notice someone posted a proof for that Fourier series thing? Heh. 
Don't know why you ask, when I said I hadn't read your proof I didn't mean to suggest I doubted it. Didn't have an actual proof in mind, seemed obvious... 
Hmm. Ok, first the continuity makes it clear that $\mu(\{t\})=0$ for all $t\ne 0$. Say $d\nu=\chi_{(0,1]}\,d\mu$. Then we want to show $\nu=0$, and we know that $\nu$ vanishes on singletons. Say $\nu([0,1])=1$. Say $\nu$ is concentrated on $E$, $m(E)=0$. 
Continuity shows that there exists a sequence $t_0<t_1\dots$ with $\nu([0,t_0])=1/2$ and $\nu((t_n,t_{n+1}])=2^{-(n+2)}$. Let $$E_n=E\cap(t_n,t_{n+1}]$$and set $$f=\sum 3^n\chi_{E_n}$$Then $f\notin L^1(\nu)$, although $f=0$ a.e.[$m$].
A: Suppose $\mu = \delta_0$. Then we have for the set $A = (0,1]$ and $B= \{0\}$ that $\mu(A) =0$ and $m(B)=0$ while $A \cup B = [0,1]$ and $A \cap B = \emptyset$. So we have property 1.
For property 2 we have $\mu ( [0,t])=1$ for all $0 \le t \le 1$, so clearly $\mu( [0,t])$ is continuous in $t$, thus also property 2 holds.
For the last property, for any $f : [0,1] \to \mathbb{R}$ we have $f(0) \in \mathbb{R}$, so $\int f \textrm{d} \mu = f(0) < \infty$, thus $f \in L^1(\mu)$.
So $\delta_0$ does satisfy all properties, and therefore is a valid counterexample.
