Is there any real number except 1 which is equal to its own irrationality measure?

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.

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.