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
{x in X | f(x) <a - (1/n) } is measurable
==> f is measurable
Is this a correct approach?
Many thanks for your feedback!
Yes, your approach is correct. Here's a more detailed explanation:
First, note that we want to prove that f is measurable, which means that the set {x in X | f(x) > a} is measurable for every real number a.
Now, for each n in the natural numbers, consider the set A_n = {x in X | f(x) < a - 1/n}. By assumption, A_n is measurable for every n.
We can rewrite {x in X | f(x) > a} as the union of A_n's:
{x in X | f(x) > a} = union of n from 1 to infinity {x in X | f(x) < a - 1/n }
To see why this is true, note that for any x such that f(x) > a, we can find a small enough n such that f(x) < a - 1/n, so x belongs to A_n.
Thus, since each A_n is measurable, their union (which is {x in X | f(x) > a}) is also measurable. Therefore, f is measurable as desired.
Why do we take union, not intersection?
We take a union because we want to show that {x in X | f(x) > a} is measurable.
Suppose we take the intersection of {x in X | f(x) < a - 1/n} for every n in the natural numbers. Then this intersection would be the set of all x in X such that f(x) is less than every real number a - 1/n, which is the same as saying that f(x) is less than or equal to a for every real number a. In other words, this intersection would always be X itself (which is certainly measurable), so it wouldn't tell us anything about whether {x in X | f(x) > a} is measurable or not.
On the other hand, taking the union of {x in X | f(x) < a - 1/n} for every n in the natural numbers gives us a set that is much closer to {x in X | f(x) > a}. Specifically, any x in {x in X | f(x) > a} belongs to {x in X | f(x) < a - 1/n} for some n (since we can always find an n such that a - 1/n < f(x)), so it belongs to the union of all these sets. Conversely, any x in the union of all these sets satisfies f(x) > a. Therefore, by showing that the union is measurable, we have shown that {x in X | f(x) > a} is measurable as well.
Your approach is close, but there is a small error in your reasoning. Let's go through it step by step.
By assuming that for each a in R, the set {x in X | f(x) <= a} is measurable, we want to prove that f is measurable.
First, let's define a new set for each n in N (the natural numbers):
B_n = {x in X | f(x) < a - 1/n}
Now, let's show that B_n is measurable for each n.
Since {x in X | f(x) <= a} is measurable, its complement {x in X | f(x) > a} is also measurable.
Now, let's consider the following set:
A_n = {x in X | f(x) >= a + 1/n}.
Note that A_n is the complement of B_n. Since both A_n and B_n are measurable, their complements should also be measurable.
So, we have:
A_n^c = {x in X | f(x) < a + 1/n} is measurable,
B_n^c = {x in X | f(x) >= a - 1/n} is measurable.
Now, let's rewrite the set {x in X | f(x) < a - 1/n} using the complement:
B_n = (B_n^c)^c = {x in X | f(x) < a - 1/n}.
Since B_n is measurable, it means that its complement B_n^c is measurable too.
Now, we have:
A_n^c = {x in X | f(x) < a + 1/n} is measurable,
B_n^c = {x in X | f(x) < a - 1/n} is measurable,
which implies that the set {x in X | f(x) < a} is measurable.
Since this holds for all a in R, it means that f is measurable.
So, in conclusion, the correct approach is to show that the measurability of the sets {x in X | f(x) <= a} implies that f is measurable.