[27] Even the most common measures can be not lower semicomputable if the parameters are not lower semicomputable. Consider a (p, 1 − p) Bernoulli process and the measure it induces on the sample space {0, 1}∞.

(a) Show that if p is a computable real number such as 1 2 or 1 3 or 1 2

√

2 or π/4 = 1 4 · 3.1415 ..., then the measure is computable.

(b) Show that if p is lower semicomputable then the measure can fail to be lower semicomputable

(c) Show that if p is a real that is not lower semicomputable, then the measure is not lower semicomputable either.

(d) Show that if p is a random real whose successive digits in its binary expansion are obtained by tosses of a fair coin, then with probability one the measure is not lower semicomputable.

Comments Hint for Item (b): If p is lower semicomputable but not computable, then 1−p is not lower semicomputable and it is the probability that the first element in the sequence is 0. Source: [P. G´acs, personal communication].