What is a Borel Measurable Function and How to Prove It?

Click For Summary
The discussion focuses on proving that the function φ(t) = t^{-1} is Borel measurable. A Borel measurable function is defined as one that pulls back sets in the Borel sigma-algebra to sets in another sigma-algebra. To establish this, it is necessary to show that the preimage of any Borel set under φ(t) is also a Borel set. The participants express confusion regarding the definition of Borel measurable functions and the steps required to demonstrate the measurability of φ(t). Clarifying these concepts is essential for completing the proof effectively.
sdf123
Messages
2
Reaction score
0

Homework Statement



Prove that the function $\phi(t)=t^{-1}$ is Borel measurable.

Homework Equations



Any measurable function into $ (\mathbb{R},\mathcal{B}(\mathbb{R}))$, where $ \mathcal{B}(\mathbb{R})$ is the Borel sigma algebra of the real numbers $ \mathbb{R}$, is called a Borel measurable function

The Attempt at a Solution



I think I need to prove that t^{-1} is a Borel set, and so prove that it is open? I am quite unclear on the actual definition of a borel measurable function, and that is perhaps my problem.
 
Physics news on Phys.org
To get your TeX to show up, enclose it in (for inline / text style) or (for equation style) tags.<br /> <br /> Now, are you familiar with the definition of a measurable function? Say you have two measurable spaces X and Y with sigma-algebras A and B, respectively. A function f:X->Y is (A-B) measurable if it pulls back sets in B to sets in A, i.e. if f<sup>-1</sup>(E) is in A whenever E is in B.<br /> <br /> A Borel measurable function f:X->Y is then an (A-B) measurable function, where B is the Borel sigma-algebra on Y. (Of course for this to make sense, Y has to be a topological space.)
 
So, in order to prove that \phi(t)=t^{-1} is Borel measurable, I need to show that if t^{-1} is a Borel sigma algebra, that {t^{-1}}^-1=t is in t, which it obviously is?
 
Question: A clock's minute hand has length 4 and its hour hand has length 3. What is the distance between the tips at the moment when it is increasing most rapidly?(Putnam Exam Question) Answer: Making assumption that both the hands moves at constant angular velocities, the answer is ## \sqrt{7} .## But don't you think this assumption is somewhat doubtful and wrong?

Similar threads

  • · Replies 8 ·
Replies
8
Views
3K
  • · Replies 2 ·
Replies
2
Views
2K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 1 ·
Replies
1
Views
2K
  • · Replies 7 ·
Replies
7
Views
7K
  • · Replies 2 ·
Replies
2
Views
2K
Replies
2
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K