Is there any real number except 1 which is equal to its own irrationality measure? Is there any real number except $1$ which is equal to its own irrationality measure? If so, then what is the cardinality of the set of all such numbers? Is the set dense on any interval? Is it measurable?
 A: Just thinking out loud, this is not my area of expertise at all but I found the question intriguing. This should be a comment but it's too long.
Using $\mu$ for the irrationality measure, and since:
$$\begin{align}\mu(x)=1&\text{if}\ x\in\mathbb{Q}\\ \mu(x)=2&\text{if}\ x\in\mathbb{A}\text{ [Roth]}\\ \mu(x)\ \overset{\underset{\mathrm{def}}{}}{=}\ \infty&\text{if}\ x\ \text{is Liouville}\\ \mu(x)\geqslant2&\text{otherwise} \end{align}$$ then $1$ is the only rational or algebraic number satisfying your condition. So any possible solution must be: (1) transcendental, (2) $>2$, and (3) not a Liouville number. Unfortunately we only know (or have upper bounds on) the irrationality measures of a very few such numbers and none of them work, and there are uncountably many transcendental numbers that are not Liouville. 
There is a construction method for numbers $x$ that have a given $\mu$, namely $$x = [\lfloor a\rfloor;\lfloor a^b\rfloor,\lfloor a^{b^2}\rfloor,\lfloor a^{b^3}\rfloor,\dots]\ |\ a>1,b=\mu-1$$ See Brisebarre, 2002. So I suppose one could set $x=\mu$ and work from there. But I don't find that anyone has done so, yet.    
A: Using the property stated in that article:
$$\mu(x)=2 + \limsup \frac{\log a_{n+1}}{\log q_n}$$
where the continued fraction expansion for $x$ is $[a_0,a_1,...]$ and the $n$th convergent is $\frac{p_n}{q_n}$.
Start with $a_0=2$ and $a_1=2$, so $q_0=1$, $q_1=2$.
Now, assume you have a continued fraction
$$\frac{p_n}{q_n}=[a_0,...,a_n]$$
Define $a_{n+1}$ to be the least integer such that $2+\frac{\log a_{n+1}}{\log q_n}>\frac{p_n}{q_n}$.
Then $x = [a_0,a_1,...] = \lim \frac{p_n}{q_n}$ will satisfy your requirement.
Just show a bound on $2+\frac{\log a_{n+1}}{\log q_n}-\frac{p_n}{q_n}$.
In particular, you can use that $\log (a_{n+1}-1)>(\log a_{n+1}) -1$ to show that if $2+\frac{\log a_{n+1}}{\log q_n}-\frac{p_n}{q_n}>\frac{1}{\log q_n}$, then $$2+\frac{\log (a_{n+1}-1)}{\log q_n}>\frac{p_n}{q_n}$$
which would violate our definition of $a_{n+1}$.  So $$\mu(x)=2+\limsup \frac{\log a_{n+1}}{\log q_n} = \lim \frac{p_n}{q_n}= x$$
So there exists such an $x$.
You can easily get uncountably many such $x$ by choosing any values $a_{2n}\in\{1,2\}$ and then choose the $a_{2n+1}$ by the above condition, again making the $\limsup$ equal to the limit of $\frac{p_n}{q_n}$.
I think the same argument can be made to show that the set is dense in $[2,\infty)$.  Basically, you can make such a $x$ starting with any finite sequence $[a_0,...,a_n]$ with $a_0\geq 2$.  Indeed, it is uncountable in any finite sub-interval $[a,b]$ with $b>a\geq 2$.
I don't think this resolves the measurability issue, contrary to my earlier claims.
It feels like $\{x:\mu(x)=x\}$ should be measurable, since it feels fairly constructive.  On the other hand, it feels like if the set $\{x:\mu(x)=x\}$ has non-zero measure, then $\{x:\mu(x)=x+\alpha\}$ should have non-zero measure, when $\alpha\in\mathbb R$, and thus we'd have an uncountable set of disjoint positive measures, which I believe is not possible.
So my guess is that the set is measurable with measure $0$.
But that is all gut, no proof.
