Let (X,@) be a measurable space.
Statement 1:
it is given that for each a in R the set
{x in X | f(x) <= a } is measurable
Statement 2: f is measurable
We need to prove that statement 1 implies 2.
My approach
Since for each a in R the set
{x in X | f(x) <= a } is measurable = intersection of of n from 1 to infinity
complement of {x in X | f(x) > a } is measurable
Since
{x in X | f(x) > a } is measurable, its complement is also measurable
==> f is measurable
Is this a correct approach?
Many thanks for your feedback!
Yes, your approach is correct. You used the fact that the sets {x in X | f(x) <= a } and {x in X | f(x) > a } are measurable for every a in R, and then used set theory to show that f(x) is measurable. Well done!
Your approach is almost correct, but there is a small mistake in your reasoning.
To prove that statement 1 implies statement 2, we can use the fact that the complement of a measurable set is also measurable.
First, let's re-write statement 1 for clarity:
For each real number a, the set {x in X | f(x) <= a} is measurable.
Now, let's assume statement 1 is true and prove statement 2:
Assume that for each real number a, the set {x in X | f(x) <= a} is measurable.
Consider any open set U in the real numbers.
We want to show that the set {x in X | f(x) < U} is measurable.
Let A_n = {x in X | f(x) <= a_n}, where a_n is a rational number such that a_n < U for all n.
Since {x in X | f(x) <= a_n} is measurable for each n, the intersection of these sets, A = ∩A_n, is also measurable.
Now, notice that the set {x in X | f(x) < U} is equal to the union of A_n for all rational numbers a_n such that a_n < U.
Therefore, the set {x in X | f(x) < U} is a countable union of measurable sets, and hence measurable.
Since this holds for any open set U, we can conclude that f is measurable.
Therefore, statement 1 implies statement 2.
Overall, your approach and reasoning are mostly correct, but you need to be careful with the complement of a set.