0
$\begingroup$

Let

$$E := \{x \in \Bbb R^5: x_1, x_2 \in \Bbb R, x_3 = x_1 + x_2, x_4 = 1, x_5 = 2\}$$

a manifold in $\Bbb R^5$. Furthermore, let $\sigma$ be the surface-measure defined by the chart $\phi: \Bbb R^2 \rightarrow E$.

Show that

$$(E, B(E), \sigma)$$

is a measure space. At first, show that $(E, B(E))$ is a measurable space.

I know that there are various theorems for showing that a function is measurable, but I have never proven that something is a measurable space. I looked up the definition for it, and it seems like that I only have to show that $B(E)$ is a $\sigma$-algebra.

But how would I have to do that? I don't know anything about $B(E)$ in this case. The $B$ might imply that it has something to do with Borel-sets, but that's all about it.

Does anyone know how to start about it?

  • 0
    I think $B(E)$ means just the Borel subsets of $E$. It's a $\sigma$-algebra by definition. What is left is to show that $\sigma$ assigns a measure to each Borel set. – 2017-02-07
  • 0
    Thanks for your answer! I don't know if this argument holds though. It's quite likely that $B(E)$ includes the Borel subsets of $E$, but nonetheless, it could be literally anything else, couldn't it? – 2017-02-07
  • 0
    In principle, yes. But if it was not specified at all, what else it could be? If it was anything, you could hardly prove it is a $\sigma$-algebra... If it's from a textbook, look for a list of symbols near the beginning or the end. – 2017-02-07
  • 0
    You're probably right, it just looks weird as an excercise. :D – 2017-02-07

0 Answers 0