Cardinality of the collection of measurable subsets of metric outer measures It is a well known fact that $\left | B_{\mathbb{R}} \right | = \left | \mathbb{R} \right |$, the standard proof involving transfinite induction. However, the next 2 statements are also true:

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*If $u$ is a metric outer measure over a metric space $X$, then every open set of the space is $u$-measurable. Then, we have as a direct corollary:

*The Borel sets of $X$ are contained in the collection of $u$-measurable sets, for every metric outer measure $u$ over $X$.

So, if we could prove the existence of a metric outer measure $u$ (over $\mathbb{R}$) such that $\left | M_u \right | = \left | \mathbb{R} \right |$, with $M_u = \left \{ u- measurable \:sets \right \}$, then the cardinality of $B_{\mathbb{R}}$ would be immediately deduced. However, this last statement seems at least a little interesting by itself. I gave some thought to it, but what i tried did not work. So, is it possible to prove the existence of such $u$?
Edit: I decided to start a bounty for this question, so i'm also posting my attempt. I understand that this problem is probably either easier than this, or it is too complex for me to understand fully, but i would still be more comfortable with an answer. I'm not exactly an expert in this field, so please forgive me if this attempt is dumb. I tried other ideas, but they ended up being false.
I tried to prove by contradiction that for every metric outer measure over $\mathbb{R}$ we have $\left | M\left ( u \right ) \right | = 2^{\left | \mathbb{R} \right |}$ (i think i had to assume AC and CH, sorry for that). So, by contradiction, suppose that $\left | M\left ( u \right ) \right | < 2^{\left | \mathbb{R} \right |}$. This means that it is not possible that every uncountable (again, i had to assume CH) set is of infinite $u$-measure, because if that were the case then we could conclude by the caratheodory condition that $\left | M\left ( u \right ) \right | = 2^{\left | \mathbb{R} \right |}$. Also (because of CH), none of this uncountable sets can be of $u$-measure $0$, because then every subset of this set would be $u$-measurable, and so (because it is uncountable and CH) we would have again $\left | M\left ( u \right ) \right | = 2^{\left | \mathbb{R} \right |}$. So then, there exists $E \subseteq \mathbb{R}$ uncountable such that $ 0 < u\left ( E \right ) < \infty$. Now, if the next statement is true, then a contradiction can be given (I don't know if it is true, but i think i could show it for several particular cases):

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*('Borelian decomposition') For very uncountable (i'm assuming CH, as stated above) subset $E \subseteq \mathbb{R}$, there exist $\left \{ E_{\alpha} \right \}_{\alpha \in \mathbb{R}} \in B_{\mathbb{R}}$ such that $\left | E_\alpha  \right | = \left | \mathbb{R} \right |$ for every $\alpha$ and $\coprod_{\alpha}^{}E_\alpha = E$ (by this i mean disjoint union) .

Now, this statement is true if we remove the condition that each $E_{\alpha}$ is borelian (to see this, take a bijection between $E$ and $\left [ 0, 1 \right ] \times \left [ 0, 1 \right ]$ and take preimages of vertical strands $\left \{ t \right \} \times \left [ 0, 1 \right ]$ for $t$ in the unit interval), but for the final contradiction i need Borelian sets, because they are $u$-measurable by hypothesis ($u$ is a metric outer measure). However, i think I could prove this ' Borelian decomposition for the following cases:

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*$E = \mathbb{R}$: using the same idea of vertical strands but this time with Peano curves.

*$E$ is a $G_{\delta}$ set (again, always uncountable): In Counterexamples in probability and real analysis, by Gary Wise, it is proven  ( problem 1.20) that for every $A \in G_{\delta}$ uncountable, there exists $B \subseteq A$ closed, nowhere dense, and such that can be mapped continously onto the unit interval (also, it is a null set in lebesgue measure, but i won't be using this property). Applying the same idea of the previous case for $B$, then we can give a 'Borelian decomposition' for $B$, and so for $A$ (if $A - B$ is uncountable, then just distribute it´s points over the borelians that constitute the Borelian decomposition of $B$)

*$E \in F_{\sigma}$ The method for this case is similar to the previous one, noticing that every closed set is a $G_{\delta}$.

*$E$ is lebesgue measurable with positive measure: This is because in this case there exists a $F \in F_{\sigma}$ and a null set $Z$ such that $E = F \coprod Z$, and the previous case gives the 'Borelian decomposition'.

*$E$ is the cantor set : I thought of this in an attempt for a counterexample, but using the Cantor -Lebesgue function i found a 'Borelian decomposition'.

I could not prove the decomposition for Lebesgue measure 0 sets, and of course not for Lebesgue non-measurable sets. But, it is still a LITTLE close. Now, to the contradiction, assuming this statement is true:
Because of the decomposition, there exist $\left \{ E_{\alpha} \right \}_{\alpha  \in \mathbb{R}}$ with the properties given in the 'Borelian decomposition' statement. Because each $E_{\alpha}$ is uncountable, it must be $u\left ( E_\alpha  \right ) > 0$, because of the first comments on my attempt (if not true for some $\alpha$, subsets of this set would be $u$-measurable). But then, consider the sets $F_n := \left \{ \alpha, u\left ( E_\alpha  \right )> 1/n \right \}$. Their union form the totality of the $\alpha \in \mathbb{R}$, but because of the decomposition, al least ONE of this subsets (name it $F_{n_0}$) must be infinite. But then, there exists $\left ( \alpha_i \right )_{i \in \mathbb{N}} \subseteq F_{n_0}$ such that $u\left ( E_{\alpha _i} \right ) > 1/n$. But, this implies that $E$ would be of infinite $u$- measure (we can decompose the measure of the disjoint union of the $E_{\alpha _i}$ in a sum of measures because these sets are borelian and thus u-measurable), a contradiction.
However, none of the cases true for the 'Borelian decomposition' necessarily work for $E$, because, $E$ can eventually be of Lebesgue measure $0$, or also not Lebesgue measurable.
Another idea i had was: suppose that $u$ is a regular metric outer measure (additional hypothesis), so the problem can be reduced to measures over $B_{\mathbb{R}}$ and induced outer measures. Then, because of the first comments for the previous attempt, $u\left ( A \right )= 0 \Rightarrow m\left ( A \right )=0$ for every subset $A$, because only countable subsets can be of $u$-measure $0$. So, if WLOG we could suppose that the induced measure over $B_{\mathbb{R}}$ is $\sigma$-finite, arguments involving (for example) Radon- Nikodym could be used (i tried but got nowhere). However, i think that proceeding this way would only give an answer for regular metric outer measures. If this is wrong, please notify me.
Any reference to this problem is also greatly appreciated.
 A: This is not a full answer, mostly a collection of remarks that might be relevant which was too large for the comments:
An obstruction this is that, if $A$ has outer measure $0$, then all the subsets of $A$, including $A$ itself, are measurable. So, if there is a set $A$ such that $\mu^*(A)=0$ and $|A|=2^{\aleph_0}$, you immediately run into trouble. Note that just having $A$ uncountable is not (at least immediately) sufficient, because you might have $2^\kappa=2^{\aleph_0}$ even though $\kappa>\aleph_0$.
Now, suppose your outer measure is well approximated by compact sets and has the following property: For every set $A$ of positive outer measure and for every $\varepsilon>0$, there exist disjoint sets $B,C\subset A$ with $0<\mu^*(B)<\varepsilon$ and  $0<\mu^*(C)<\varepsilon$. Then you can run the argument used in $\mathbb{R}$ to construct the ordinary Cantor set. Compactness takes care of the fact that infinite intersections of (certain) decreasing sets are non-empty. You can also get the desired conclusion when the ambient metric space is complete and the sets $B$ and $C$ above can be taken with arbitrarily small diameter.

This is less relevant to your exact question, but one can prove that the collection of Borel sets has the size of the continuum without using any transfinite induction/recursion. The key ingredient is:

Theorem: Let $B\subset \mathbb{R}$ be a Borel set. Then there is a continuous function $f:\mathcal{N}\to \mathbb{R}$, where $\mathcal{N}$ is the space of (countable) sequences of natural numbers (usually called Baire space), such that $f[\mathcal{N}]=B$.

For a proof, see Kechris' Classical Descriptive Set Theory or Srivastava's A Course on Borel Sets. In fact, $\mathbb{R}$ can be replaced by any Polish space. A Polish space is a topological space which is separable and completely metrizable. The proofs in the above books should be accessible to anyone with some knowledge of general topology (and a tiny bit of real analysis).
The bound $|\mathcal{B}_\mathbb{R}|\le 2^{\aleph_0}$ now follows immediately, because there are only continuum many continuous functions  $\mathcal{N}\to\mathbb{R}$.
A: We can proceed continue from Reveillark’s answer as follows.

An obstruction this is that, if $A$ has outer measure $0$, then all the subsets of $A$, including $A$ itself, are measurable. So, if there is a set $A$ such that $\mu^*(A)=0$ and $|A|=2^{\aleph_0}$, you immediately run into trouble.

Let $d$ be the given metric on a set $\Bbb R$ and $u$ be the given finite outer measure on the space $X=(\Bbb R, d)$ (the case when $u$ is $\sigma$-finite can be reduced to this, because, by the Kőnig Theorem, the cofinality of $\frak c>\omega$, that is a set of size $\frak c$ cannot be a union of countably many sets of a smaller size). According to Subsection 17.6 from Kechris’ book (Springer 1995 edition), a standard result in measure theory asserts that $u$ is a metric outer measure on a metric space $X$ iff every Borel set in $X$ is $u$-measurable, so in this case the restriction $u’$ of $u$ onto the $\sigma$-algebra ${\bf B}(X)$ of Borel subsets of $X$ is a Borel measure.
If the metric space $X$ is non-separable then there exists $\varepsilon>0$ and an uncountable subset $B$ of $X$ such that $d(x,y)>\varepsilon$ for each distinct points $x,y\in X$. It follows that each subset of $B$ is closed.  Let $\{B_\alpha:\alpha\in \mathcal I\}$ be a partition of $B$ into uncountable many uncountable subsets. The finiteness of $u$ implies that for each $n>0$ a set ${\alpha\in I:u(B_\alpha)}>1/n$ is finite. Since the set $\mathcal I$ is uncountable, it contains and index $\alpha$ such that $u(B_\alpha)=0$. Since $B_\alpha$ has at least $2^{\omega_1}$ distinct subsets, we have a contradiction provided $2^{\omega_1}>\frak c$.
I expect that in a general case we can construct a required set $A$ as follows. Suppose that  $X$ contains a Borel set $B$ such that $|B\setminus C|=\frak c$ for each $F_\sigma$-set $C$ of $X$. I guess this always holds and I asked at MSE a separate question about this. By Theorem 17.10 from the above book, $u’$ is regular, so there exists an $F_\sigma$ subset $C$ of $B$ such that $u’(C)=u’(B)=u(B)$. Put $A=C\setminus B$. Then $|A|=\frak c$ and $u(A)=0$.
