1

This is regarding Theorem 1.8 on page 11 of Rudin's Real and Complex Analysis, we have a measurable space $X$ and two real valued functions on it $u$ and $v$. We then have a continuous mapping $\Phi$ from the real plane into a topological space $Y$. We define $f(x)=(u(x),y(x))$ and $h=\Phi\circ f$, and from here we need to show that $h$ is measurable and we know that it suffices to show that $f$ is measurable.

We then take a rectangle $R$ in the plane with sides parallel to the axes, which we write as the Cartesian product of $I_1$ and $I_2$ which are two segments, presumably two adjacent sides of the rectangle. We then say that: $$f^{-1}(R)=u^{-1}(I_1)\cap v^{-1}(I_2) \tag{1},$$ and this is what's confusing me. $I_1$ and $I_2$ are essentially just lines in the plane, why would the intersection of their preimage be equal to the preimage of the entire rectangle and not just two "lines" in $X$?

Disclaimer: I am a physicist not a mathematician please have mercy on my soul if this is a trivial error on my part.

Charlie
  • 685
  • It might help to (initially, incorrectly) think of $f(x)$ as a "line" since it's just parametrized by $x$. You're then asking what values of $x$ makes this "line" fall into the rectangle $R$. – Calvin Lin Nov 07 '20 at 14:45

1 Answers1

2

As far as I understand, we have $R=I_1\times I_2$.

Now we want to show $f^{-1}(R)=u^{-1}(I_1)\cap v^{-1}(I_2)$.

This is a straight forward proof:

$f^{-1}(R)=\{x\in X: f(x)\in R\}=\{x\in X: (u(x),v(x))\in I_1\times I_2\}=\{x\in X: u(x)\in I_1~\text{and}~ v(x)\in I_2\}$

$=\{x\in X: u(x)\in I_1\}\cap \{x\in X: v(x)\in I_2\}=u^{-1}(I_1)\cap v^{-1}(I_2)$

Cornman
  • 11,065
  • 4
  • 30
  • 57
  • So are $I_1$ and $I_2$ "lines" in the plane? It's still strange to me if that's the case that their preimage is more than just a "line" in $X$? – Charlie Nov 07 '20 at 14:45
  • The cartesian product of two sets is defined as $A\times B={(a,b): a\in A~\text{and}~ b\in B}$. So for example take the $A=B=[0,1]$. The intervall. This is the product of 2 lines, but the point (1/2, 1/2) is in the cartesian product, as $1/2\in A$ and $1/2\in B$. If you draw a corresponding picture, then you see that this poin lies in the square. Do you see it? – Cornman Nov 07 '20 at 14:52
  • Would it be correct to say that all points inside $I_1\times \Bbb R$ are in the image of $u(u^{-1}(I_1))$? In other words $I_1$ is essentially the entire "strip" through $\Bbb R^2$ such that the values lie within some interval on one of the axes? – Charlie Nov 07 '20 at 15:06
  • Do you mean $u^{-1}(u(I_1))$? Other way around it makes no sense, as $I_1$ is not a subset of $Y$, but of $X$. You are correct that $I_1\times\mathbb{R}$ can be visualised as a big "strip" through $\mathbb{R}^2$. But $I_1\times\mathbb{R}$ has nothing to do with the set $u(u^{-1}(I_1))$ or $u^{-1}(u(I_1))$, where the first one makes no sense, as I said and for the second it should be $u^{-1}(u(I_1))\subseteq I_1$. The problem is that $I_1\subseteq X$, which has nothing to do with a cartesian product of a set with $\mathbb{R}$. So you are confusing some sets here, I guess. – Cornman Nov 07 '20 at 15:12
  • Oh wait $I_1$ and $I_2$ are in $\Bbb R^2$ are they not? $u:X\rightarrow R$ since it's a real valued function so $u(I_1)$ doesn't make sense no? I think either way I've understood what you mean so thank you very much. – Charlie Nov 07 '20 at 15:19
  • 1
    I am not quite sure, as there are some undefined sets and functions. So I do not know what exactly $X$ and $Y$ is. The functions u and v are supposed to be real-valued. So $u: X\to\mathbb{R}$ and $v: X\to\mathbb{R}$. The plane is supposed to be the $\mathbb{R}^2$ and a rectangle in a plane has sides in $\mathbb{R}$. And yes, $u(I_1)$ does not make sense, but $u^{-1}(I_1)$ does. Sorry, I seem to confused it in the comment above. – Cornman Nov 07 '20 at 15:26
  • I see, thanks for your help I understand now :) – Charlie Nov 07 '20 at 15:28
  • You are welcome. – Cornman Nov 07 '20 at 15:30