Are there some measures of information content of Bayesian evidence? Let us have a binary random variable $X$ with values $a,b$ and a prior distribution $P(a), P(b)$. Now, let us suppose that we learn a piece of evidence $E$. $E$ is a random variable with two values, $A, B$, where
$$
P(E = A\mid a) = \alpha, P(E = B\mid a) = 1 - \alpha \\
P(E = A\mid b) = \beta, P(E = B\mid b) = 1 - \beta
$$
If we want to determine $X$, we should seek out the most informative evidence possible. It is clear, that any evidence where $\alpha = \beta$ is equally useless (contains no information). On the other hand, $\alpha = 1, \beta = 0$ is maximally useful (contains the maximum amount of information).
I was looking for ways to formalize this "information content" or "discriminative value" of evidence in some nice way that would generalize intuitively to variables with $n$ possible values.
Since entropy is the measure of information content of a random variable, I tried to see what is the entropy of $E$:
$$
-(P(a)\alpha + P(b)\beta) \log (P(a)\alpha + P(b)\beta) -\\
(P(a)(1-\alpha) + P(b)(1-\beta)) \log (P(a)(1-\alpha) + P(b)(1-\beta))
$$
Unfortunately this does not express the measure of information in a way that I would expect. If we plug the useless zero information case where $\alpha = \beta$ into the expression we get:
$$
-\alpha \log (\alpha) - (1-\alpha) \log (1-\alpha)
$$
which is not constant as we would expect. Furthermore, $\alpha = 1, \beta = 0$ reduces to the entropy of $X$, which is also not useful. 
What are some other measures of information that measure information in the sense of discriminative value?
 A: Your problem is studied in the field of Bayesian experimental design. The question that experimental design faces is, which measure should I make in order to maximize the information gain? Or put in another way, which measure of $E$ is going to allow me to discriminate the most between different values of the parameter $X$?
A possible way to measure the information gain is the Kullback-Leibler divergence (https://en.wikipedia.org/wiki/Kullback%E2%80%93Leibler_divergence#Bayesian_updating). Its form is very close to the expression of the entropy that you used.
$$ U(\boldsymbol{y}) = \int d^{n}\theta P(\boldsymbol{\theta} | \boldsymbol{y}) \textrm{log}(\frac{P(\boldsymbol{\theta} | \boldsymbol{y})}{P(\boldsymbol{\theta})})$$
which in your case can be written as 
$$ U(X) = \sum_{i} P(E=E_i|x)\textrm{log}(\frac{P(E=E_i|x)}{P(E=E_i)})  = P(E=A|x)\textrm{log}(\frac{P(E=A|x)}{P(E=A)}) + P(E=B|x)\textrm{log}(\frac{P(E=B|x)}{P(E=B)})$$
More explicitely,
$$ U(x=a) = P(E=A|a)\textrm{log}(\frac{P(E=A|a)}{P(A)}) + P(E=B|a)\textrm{log}(\frac{P(E=B|a)}{P(B)})$$
$$ U(x=b) = P(E=A|b)\textrm{log}(\frac{P(E=A|b)}{P(A)}) + P(E=B|b)\textrm{log}(\frac{P(E=B|b)}{P(B)})$$
Substituting the values that you provide and taking into account that $P(E=A)=\alpha P(a) + \beta P(b)$ and that $P(E=B)=(1-\alpha) P(a) + (1-\beta) P(b)$, we get
$$ U(x=a) = \alpha \textrm{log}(\frac{\alpha}{\alpha P(a)+ \beta P(b)}) + (1-\alpha) \textrm{log}(\frac{1-\alpha}{(1-\alpha) P(a)+ (1-\beta) P(b)})$$
$$ U(x=a) = \beta \textrm{log}(\frac{\beta}{\alpha P(a)+ \beta P(b)}) + (1-\beta) \textrm{log}(\frac{1-\beta}{(1-\alpha) P(a)+ (1-\beta) P(b)})$$
Now, if you set $\alpha=\beta$ and take into account that $P(a)+P(b)=1$, you will see that in fact $U(x=a)=U(x=b)=0$, as expected. On the other hand, if $\alpha=1$ and $\beta=0$, $U(x=a)=\textrm{log}(1/P(a))$ and $U(x=b)=\textrm{log}(1/P(b))$ so the information gain still depends on the prior probabilities.
