You are correct that without the axiom of choice $2^{\aleph_0}\newcommand{\CH}{\mathsf{CH}}$ may not be an $\aleph$. Therefore the continuum hypothesis split into two inequivalent statements:
- $(\CH_1)$ $\aleph_0<\mathfrak p\leq2^{\aleph_0}\rightarrow2^{\aleph_0}=\frak p$.
- $(\CH_2)$ $\aleph_1=2^{\aleph_0}$.
Whereas the second variant implies that the continuum is well-ordered, the first one does not.
You suggested a third variant:
- $(\CH_3)$ $\aleph_0<\mathfrak b\rightarrow 2^{\aleph_0}\leq\mathfrak b$.
Let's see why $\CH_3\implies\CH_2\implies\CH_1$, and that none of the implications are reversible.
Note that if we assume $\CH_3$, then it has to be that $2^{\aleph_0}\leq\aleph_1$ and therefore must be equal to $\aleph_1$. If we assume that $\CH_2$ holds, then every cardinal less or equal to the continuum is finite or an $\aleph$, so $\CH_1$ holds as well.
On the other hand, there are models of $\sf ZF+\lnot AC$, such that $\CH_1$ holds and $\CH_2$ fails. For example, Solovay's model in which all sets are Lebesgue measurable is such model.
But $\CH_2$ does not imply $\CH_3$ either, because it is consistent that $2^{\aleph_0}=\aleph_1$, and there is some infinite Dedekind-finite set $X$, that is to say $\aleph_0\nleq |X|$. Therefore we have that $\aleph_0<|X|+\aleph_0$. Assuming $\CH_3$ would mean that if $X$ is infinite, then either $\aleph_0=|X|$ or $2^{\aleph_0}\leq|X|$. This is certainly false for infinite Dedekind-finite sets (one can make things stronger, and use sets that have no subset of size $\aleph_1$, while being Dedekind-infinite).
One can also think of the continuum hypothesis as a statement saying that the continuum is a certain kind of successor to $\aleph_0$. As luck would have it, there are $3$ types of successorship between cardinals in models of $\sf ZF$, and you can find the definitions in my answer here.
It is easy to see that $\CH_1$ states "$2^{\aleph_0}$ is a $1$-successor or $3$-successor of $\aleph_0$", and $\CH_3$ states that "$2^{\aleph_0}$ is a $2$-successor of $\aleph_0$" -- while not explicitly, it follows from the fact that I used to prove $\CH_3\implies\CH_2$.
So where does $\CH_2$ gets here? It doesn't exactly get here. Where $\CH_1$ and $\CH_3$ are statements about all cardinals, $\CH_2$ is a statement only about the cardinality of the continuum and $\aleph_1$. So in order to subsume it into the $i$-successor classification we need to add an assumption on the cardinals in the universe, for example every cardinal is comparable with $\aleph_1$ (which is really the statement "$\aleph_1$ is a $2$-successor of $\aleph_0$").
All in all, the continuum hypothesis can be phrased and stated in many different ways and not all of them are going to be equivalent in $\sf ZF$, or even in slightly stronger theories (e.g. $\sf ZF+AC_\omega$).
Without the axiom of choice we can have two notions of ordering on the cardinals, $\leq$ which is defined by injections and $\leq^*$ which is defined by surjections, that is to say, $A\leq^* B$ if there is a surjection from $B$ onto $A$, or if $A$ is empty. These notions are clearly the same when assuming the axiom of choice but often become different without it (often because we do not know if the equivalence of the two orders imply the axiom of choice, although evidence suggest it should -- all the models we know violate this).
So we can formulate $\CH$ in a few other ways. An important fact is that $\aleph_1\leq^*2^{\aleph_0}$ in $\sf ZF$, so we may formulate $\CH_4$ as $\aleph_2\not\leq^*2^{\aleph_0}$. This formulation fails in some models while $\CH_1$ holds, e.g. in models of the axiom of determinacy, as mentioned by Andres Caicedo in the comments.
On the other hand, it is quite easy to come up with models where $\CH_4$ holds, but all three formulations above fail. For example the first Cohen model has this property.
All in all, there are many many many ways to formulate $\CH$ in $\sf ZF$, which can end up being inequivalent without some form of the axiom of choice. I believe that the correct way is $\CH_1$, as it captures the essence of Cantor's question.
Interesting links:
- What's the difference between saying that there is no cardinal between $\aleph_0$ and $\aleph_1$ as opposed to saying that...
- Relationship between Continuum Hypothesis and Special Aleph Hypothesis under ZF
Of course assuming CH they can. But if $\aleph_1 < 2^{\aleph_0}$ then there is no measurable set of cardinality $\aleph_1$ having positive Lebesgue measure, so the answer to the question in the title is no. We can prove the following in ZFC without requiring additional axioms.
Proposition (ZFC). Every uncountable Borel subset of $\mathbb{R}$ has cardinality $2^{\aleph_0}$.
This is a corollary of the Perfect Set Theorem (every uncountable Borel set contains a perfect set, i.e. a set with no isolated points, and any such set has cardinality at least $2^{\aleph_0}$). For a proof, see Theorem 13.6 of Kechris's Classical Descriptive Set Theory.
Corollary (ZFC). Every Lebesgue measurable subset of $\mathbb{R}$ with cardinality less than $2^{\aleph_0}$ has Lebesgue measure zero.
Proof. Every Lebesgue measurable set $E$ can be written $E = B \cup N$ where $B$ is Borel and $m(N) = 0$; in particular $m(E) = m(B)$. But if $|E| < 2^{\aleph_0}$ then $|B| < 2^{\aleph_0}$, so by the previous proposition $B$ is countable and hence $m(B) = 0$.
Best Answer
The basic strategy of an earlier answer of Asaf Karagila's can be used here too to show that there is no formula $\varphi$ which $\mathsf{ZFC}$ proves defines a set of reals of size $\aleph_1$. (Really, one proves: if $\mathsf{ZFC}\vdash\{x\in\mathbb{R}:\varphi(x)\}$ is uncountable then there is a forcing extension in which that set has size continuum.) The details are a bit technical, but I suspect this is folklore.
This, combined with the usual descriptive set theoretic obstacles (assuming $\neg\mathsf{CH}$ there is no Borel set of size $\aleph_1$, nor a projective set of size $\aleph_1$ if there are also large cardinals), gives a pretty strong negative answer.
Beyond this, you will need to make more precise your notion of "explicit construction."