Does every normal number have irrationality measure $2$? A normal number is a number whose digit expansion in any base is "uniform" in the sense that all finite digit strings occur with the "statistically expected" frequency.

I read a sentence somewhere which could be understood as implying that every normal number had irrationality measure $2$.

Is it known if all normal numbers have irrationality measure (approximation exponent) $2$?

Also, are there some non-normal irrational numbers whose irrationality measures are known?

Also, are there some non-normal numbers with irrationality measure $2$ known?
normal $\; \not\Rightarrow \;$ irrationality measure $2$

There exist numbers that are normal with irrationality measure $&gt;2$. In fact, there exist normal numbers (meaning normal with respect to every base) that have irrationality measure $\infty.$ This is Theorem 2 in Bugeaud [1] (2002). For related results, see [2] and [3].

irrationality measure $2$ $\; \not\Rightarrow \;$ normal

There exist numbers with irrationality measure $2$ that are not normal. In fact, there exist numbers with irrationality measure $2$ that fail to be simply normal in every base. Note that this is stronger than failing to be simply normal in base $10,$ which in turn is stronger than failing to be normal in base $10,$ which in turn is stronger than "normal" in the sense that you are asking about. Shallit [4] (1979) showed that the continued fraction expansion of

$$\sum_{n=0}^\infty 2^{-2^{n}}$$

has bounded partial quotients (see Theorem 3 on p. 213 for meaning of $B(u,\infty)$ and Theorem 9 on p. 216 for the result), and thus has irrationality measure $2.$ [Indeed, for this we only need the $n$th partial quotient to be bounded by a linear function of $n.$ See Robert Israel's answer to the math overflow question Numbers with known irrationality measures.] However, it is clear that this number is not simply normal in any base. Indeed, in this number's expansion in any base, the proportion of $0$'s approaches $1,$ and hence the proportions for each of the other digits approaches $0.$

[1] Yann Bugeaud, Nombres de Liouville et nombres normaux [Liouville numbers and normal numbers], Comptes Rendus Hebdomadaires des Séances de l'Académie des Sciences 335 #2 (2002), 117-120.

[2] Verónica Becher, Pablo Heiber and Theodore A. Slaman, A computable absolutely normal Liouville number, preprint, 30 January 2014, 14 pages.

[3] Satyadev Nandakumar and Santhosh Kumar Vangapelli, Normality and finite-state dimension of Liouville numbers, arXiv:1204.4104v2, 21 January 2014.

[4] Jeffrey Outlaw Shallit, Simple continued fractions for some irrational numbers, Journal of Number Theory 11 #2 (April 1979), 209-217.

(UPDATE, 41 MONTHS LATER) A few days ago I happened to come across [5] below, which might be of interest to those finding this web page from an internet search for some known relations between normal numbers and Liouville numbers.

[5] Richard George Stoneham, A general arithmetic construction of transcendental non-Liouville normal numbers from rational fractions, Acta Arithmetica 16 #3 (1970), 239-253.
The Dirichlet transcedentals, which are of the form
$$
a=\sum_{n\in\mathbb N} a_n 10^{-n!}, \quad a_n\in\{1,2\},
$$
are as many the real numbers. 

The irrationality measure of these numbers is infinite, and hence they are not normal!