Induced Bernoulli measures Given a sequence of Bernoulli trials with paramater $p$. We have by Kolmogorov's extension theorem, a probability measure $P^\mathbb{N}$ on the space of binary sequences. Now letting $P$ denote the measure on $[0,1]$ that corresponds to $P^\mathbb{N}$ if we represent numbers in their binary representation. I would like to prove that $P((0,x])$ is continuous. Furthermore, since each $0<p<1$ induces a new measure $P_p$. Is it always true that these measures are mutually singular?
For the latter, I need to construct disjoint sets $A_p$ such that $P_p(A_p)=1$ and $P_p(A_q)=0$ for $p\ne q$. But I am unsure how to do this. I believe LLN is useful but we have uncountable many measures.
I have got $P((0,x])=\sum_n \frac{x_n}{2^n}$ where $x\in[0,1]$ is represented in binary by $(x_n)_n$
 A: Denote by $f:[0,1] → \mathbb R$ the function in question:
$$f (x) := \mathbb P \big((0,x]\big).$$
Firstly, I am not convinced your formula at the end of the question holds: i.e.,
$$f(x) = \sum_{n=1}^∞ \frac{x_n}{2^n},$$
since the right hand side equals $x$! And this is far too rigid: indeed, $f(1/2) = p$ is not necessarily equal to $1/2$.
Part 1
Let's take another approach:
For simplicity, fix $x \in (0,1)$; the endpoints can be dealt with similarly.
The important thing is that $f$ is monotone increasing; therefore both the limit from below
$$f(x^-) := \lim_{\epsilon → 0^+} f(x - \epsilon)$$
and the limit from above
$$f(x^+) := \lim_{\epsilon → 0^+} f(x + \epsilon)$$
exist (this can be deduced from the completeness axiom for monotone sequences).
Moreover, $f$ is discontinuous at $x$ if and only if these limits are distinct: $f(x^-) < f(x^+)$; this is a a so-called jump discontinuity.
To delve deeper:
$$\begin{align} f(x^-) := \lim_{\epsilon→ 0^+} f(x - \epsilon) &= \lim_{\epsilon → 0^+} \mathbb P \big((0,x-\epsilon]\big)  \\&\stackrel\ast= \phantom{\lim_{\epsilon → 0^+}{}}\mathbb P \left(\bigcup_{\epsilon > 0}\; (0,x-\epsilon]\right)\\&= \phantom{\lim_{\epsilon → 0^+}{}} \mathbb P \big((0,x)\big),\end{align}$$
where the equality marked by $\ast$ is continuity from below. Similarly,
$$\begin{align} f(x^+) := \lim_{\epsilon→ 0^+} f(x + \epsilon) &= \lim_{\epsilon → 0^+} \mathbb P \big((0,x+\epsilon]\big)  \\&\stackrel\ast= \phantom{\lim_{\epsilon → 0^+}{}}\mathbb P \left(\bigcap_{\epsilon > 0}\; (0,x+\epsilon]\right)\\&= \phantom{\lim_{\epsilon → 0^+}{}} \mathbb P \big((0,x]\big)\\&=:\; \phantom{\lim_{\epsilon → 0^+}{}} f(x). \end{align}$$
Therefore, these two one-sided limits are equal (to $f(x)$) if and only if
$$\mathbb P (\{x\}) = 0.$$
Let $(a_n)_{n=1}^∞ \in \{0,1\}^{\mathbb N}$. Then
$$\sum_{n = 1}^∞ \frac{a_n}{2^n} = \sum_{n = 1}^∞ \frac{x_n}{2^n}$$
implies either that

*

*$a_n = x_n$ for all $n \in \mathbb N$; or

*$x = j/2^N$ for some $N\in \mathbb N$ and odd number $j$:
$$ x = \sum_{n = 1}^N \frac{x_n}{2^n},\quad x_N=1\quad \text{and}\quad a_n = \begin{cases} x_n, & n < N, \\ 0, & n = N, \\ 1, & n > N \end{cases}$$
(i.e., just like $1 = 0.9999\ldots$, but in binary).

Both of these are probability $0$ events, so you're done: the left and right limits exist and are equal to $f(x)$, so the $f$ is continuous at $x$.
Part 2
For $p \in [0,1]$, consider the set of all sequences where $1$s appear with asymptotic frequency $p$, $A_p$:
$$A_p := \left\{(a_n)_{n=1}^∞ \in \{0,1\}^{\mathbb N}\, \bigg|\, \lim_{N → ∞}\underbrace{\vphantom{\bigg|}\frac{\#\big\{n \in \{1,2,\ldots, N\} : a_n = 1\big\}}{N}}_{S_N}  = p \right\}$$

Claim: For all $p \in [0,1]$, $$\mathbb P_p (A_p) = 1.$$
Proof: Fix $p$, And let $(a_n)_{n\geq 1}$ denote a sequence of iid Bernoulli trials (i.e. a random element in the space). Then
$$S_N = \frac 1N \sum_{n=1}^N \boldsymbol 1 _{a_n=1} = \frac 1N \sum_{n=1}^N a_n$$
is the average of the first $N$ trials. Therefore by the Law of Large Numbers, $S_N$ converges almost surely to the expected value of one trial, $p$. In other words, $A_p$ has full measure, $\mathbb P (A_p) = 1$.

Claim: for all $p\neq q$, $A_p$ and $A_q$ are disjoint.
Proof: $(a_n) \in A_p \implies S_N → p \implies S_N \not → q \implies (a_n)\not \in A_q$, and vice versa.

Claim. Let $p\neq q$. then $\mathbb P (A_q) = 0$
Proof: Since $\mathbb P (A_p) = 1$ and $A_p$, $A_q$ are disjoint:
$$ 1 + \mathbb P_p(A_q) = \mathbb P_p (A_p) + \mathbb P_p (A_q) = \mathbb P_p (A_p \sqcup A_q)) \leq 1,$$
i.e. $\mathbb P_p(A_q) \leq 0$.
