Show that E is measurable if and only if kE is measurable.
We need to show two things:
1. If E is measurable, then kE is measurable.
2. If kE is measurable, then E is measurable.
Proof of 1:
Assume E is measurable. We want to show that kE is measurable. Let A be an arbitrary subset of the real line. We have to show that kE intersects A in a measurable set if and only if E intersects k^{-1}(A) in a measurable set.
Let's first assume that kE intersects A in a measurable set. Then there exists a Borel set B such that kE ∩ A = B. Since E and k^{-1}(A) are related by the equation E = (1/k)kE and k^{-1}(A) = k^{-1}(1/k)kE, we have:
E ∩ k^{-1}(A) = (1/k)kE ∩ k^{-1}(1/k)kE = (1/k)kE ∩ E = (1/k)B
Since B is Borel, (1/k)B is also Borel, which means E intersects k^{-1}(A) in a measurable set. Therefore, kE is measurable.
Proof of 2:
Assume kE is measurable. We want to show that E is measurable. Let A be an arbitrary subset of the real line. We have to show that E intersects A in a measurable set if and only if kE intersects k(A) in a measurable set.
Let's first assume that E intersects A in a measurable set. Then there exists a Borel set B such that E ∩ A = B. Since kE and k(A) are related by the equation kE = k(k^{-1}(A) ∩ E), we have:
kE ∩ k(A) = k(k^{-1}(A) ∩ E) ∩ k(A) = k((k^{-1}(A) ∩ E) ∩ k^{-1}(k(A))) = k((k^{-1}(A) ∩ E) ∩ A)
Since B is Borel, (k^{-1}(A) ∩ E) ∩ A is also Borel, which means kE intersects k(A) in a measurable set. Therefore, E is measurable.
Hence, we have shown that E is measurable if and only if kE is measurable.
To show that E is measurable if and only if kE is measurable, we need to prove two statements:
1. If E is measurable, then kE is measurable.
2. If kE is measurable, then E is measurable.
Let's start with the first statement:
1. If E is measurable, then kE is measurable.
Assume that E is measurable. We want to show that kE is measurable.
Since E is measurable, its outer measure is defined as:
m*(E) = inf{∑(i=1 to ∞) |Q_i| : E ⊆ ∪(i=1 to ∞) Q_i, and Q_i is a closed cube}
where |Q_i| denotes the volume (or measure) of the closed cube Q_i.
Now, consider the set kE. We want to show that its outer measure is defined as:
m*(kE) = inf{∑(i=1 to ∞) |Q_i| : kE ⊆ ∪(i=1 to ∞) Q_i, and Q_i is a closed cube}
By definition, kE = {ke : e ∈ E}, where k is a constant.
Now, consider the closed cubes Q_i that cover kE. Each closed cube Q_i corresponds to a cube Q'_i that covers E, with the size scaled (by a factor of 1/k) from Q_i.
Since E is measurable, we have:
m*(E) = inf{∑(i=1 to ∞) |Q'_i| : E ⊆ ∪(i=1 to ∞) Q'_i, and Q'_i is a closed cube}
Now, let Q_i = kQ'_i, where Q_i is a closed cube that covers kE and Q'_i is a closed cube that covers E.
We can write:
m*(kE) = inf{∑(i=1 to ∞) |kQ'_i| : kE ⊆ ∪(i=1 to ∞) kQ'_i, and kQ'_i is a closed cube}
= k * inf{∑(i=1 to ∞) |Q'_i| : E ⊆ ∪(i=1 to ∞) Q'_i, and Q'_i is a closed cube}
= k * m*(E)
Since k is a constant, k * m*(E) is still a finite value, which implies that kE is measurable.
Therefore, if E is measurable, then kE is measurable.
Now, let's move on to the second statement:
2. If kE is measurable, then E is measurable.
Assume that kE is measurable. We want to show that E is measurable.
Let Q_i be closed cubes that cover E.
Then, kQ_i will cover kE.
Since kE is measurable, we have:
m*(kE) = inf{∑(i=1 to ∞) |kQ_i| : kE ⊆ ∪(i=1 to ∞) kQ_i, and kQ_i is a closed cube}
Since E ⊆ kE, we can write:
m*(kE) >= inf{∑(i=1 to ∞) |kQ_i| : E ⊆ ∪(i=1 to ∞) kQ_i, and kQ_i is a closed cube}
Now, divide both sides by k (which is non-zero):
m*(E) >= inf{∑(i=1 to ∞) |Q_i| : E ⊆ ∪(i=1 to ∞) Q_i, and Q_i is a closed cube}
This inequality implies that E is measurable.
Therefore, if kE is measurable, then E is measurable.
Combining both statements, we conclude that E is measurable if and only if kE is measurable.