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Let

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

be a manifold. Furthermore, let

$$\phi: \Bbb R^2 \rightarrow E$$

be a chart and $\sigma$ the surface-measure. You are allowed to assume that $(E, B(E))$ is a measurable space.

Show that there exists a $c \in \Bbb R$ such that

$$\sigma(A) = c\lambda^2(\phi^{-1}(A))$$

and conclude that $\sigma$ is a measure.

Since the surface measure $\sigma$ is independent from the chart we choose, we define a new chart $\phi: \Bbb R^2 \rightarrow E$ by

$$\phi(x_1, x_2) := \begin{pmatrix} x_1 \\ x_2 \\ x_1 + x_2 \\ 1 \\ 2 \\ \end{pmatrix}.$$

By definition of the surface measure and the integral on manifolds, it follows that (for $X_A$ being the indicator function)

$$\sigma(A) = \int_A 1 d\sigma = \int_E X_A d\sigma = \int_{\Bbb R^2} X_A(\phi(x_1, x_2)) \sqrt {g} \ d\lambda^2(x, y).$$

$g$ refers to the determinant of the gramian matrix of $M$ with respect to $\phi$. A little bit of calculating gives $g = 4$, hence $\sqrt g = 2$, which is a constant of course. Hence, the above integral can be written as

$$2 \int_{\Bbb R^2} X_A(\phi(x, y)) d\lambda^2(x, y) = 2 \int_{\Bbb R^2} X_{\phi^-1(A)} d\lambda^2(x, y).$$

For general measures $\theta$ on a set $\Omega$, we know the identity

$$\theta(A) = \int_{\Omega} X_A d\theta.$$

Hence, the above integral is identical to

$$2 \lambda^2(\phi^{-1}(A))$$

with $c := 2$ being our desired constant,

so

$$\sigma(A) = c \lambda^2(\phi^{-1}(A)).$$

This is identical with the definiton of the Pushforward measure, which makes $\sigma(A)$ a meausure too.

Is that a decent solution?

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    If somebody answers "yes, I have checked it and this is a decent solution" will you accept that answer and award it the bounty? – 2017-02-14
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    If that person checked it carefully and didn't find any mistakes - sure! The reason for the bounty is just that it is an old exam question that might appear similar on my exam on monday, so I want to make sure that I definitively got it right. ;) – 2017-02-14

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There are two problems with your solution: a minor one and a big one.

The minor one is that the first fundamental form (i.e. the Riemannian metric) of $E$ is, in your chosen coordinates, $\begin{pmatrix} 2 & 1 \\ 1 & 2 \end{pmatrix}$, therefore its determinant is $3$, not $4$.

The big one is that the problem requires you to show that equality for any global chart $\phi$, whereas you only show it for the specific chart that you construct.

There is a third problem, too, that you might not be responsible of, though: if $\sigma$ is said to be the surface-measure of $E$, why are you required to conclude that it is a measure, anymore? Alternatively, what is a "surface-measure" if not a measure? (I do not know the construction and definition of $\sigma$ that you are working with, there are several of them possible - all of them equivalent in the end.)

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    Thanks for your answer! Maybe I should specify: I am supposed to show that $\sigma$ is a measure on $B(E)$, not a measure "in general". Does this make a difference? Because we had an assignment with a different manifold that worked the same: defining a chart, applying the identities, making the conclusion with the pushforward measure and hence, conclude that $\sigma$ is a measure on (in this case) $B(M)$. – 2017-02-14
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    @Julian: There are no "measures in general": every measure must live on some $\sigma$-ring - in this case, it's $B(E)$. So no, your comment makes no difference. – 2017-02-14
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    Well, okay. In our assignment, the manifold was $M := S^1 \setminus \{(1, 0)\}, S^1 := \{x \in \Bbb R^2 : ||x|| = 1\}$. Our tutor defined the chart $\phi: (0, 2\pi) \rightarrow M$ with $\phi(\alpha) = (\cos \alpha, \sin \alpha)$. Then he proceeded the same way I did above. Is this wrong too? – 2017-02-14
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    @Julian: I cannot comment on work that I can't see for myself. Maybe the original statement was "construct a convenient chart and use it to show that bla-bla-bla"? – 2017-02-14
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    No, we had the manifold $M$, the chart $\phi: A \rightarrow M, A \subset \Bbb R$, the surface measure $\sigma$ that is assigned by this chart, and it we had to show that $(M, B(M), \sigma)$ is a measure space, so one had to show that $B(M)$ is a $\sigma$-algebra and that $\sigma$ is a measure on $B(M)$. – 2017-02-14
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    But as I mentioned in the beginning, the surface measure is independent from the chart we choose, so wouldn't this mean that I showed it for any arbitrary (global) chart? – 2017-02-14
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    I talked to my tutor again and he told me that my solution was correct. Of course my chart is a global chart since it maps into every point of the manifold, and as I said, the surface measure / the value of the integral is independend from the map we choose, so this is an "arbitrary global map". I would like to give you a source for this, but I got my information from a German book. – 2017-02-14
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    @Julian: I am afraid that your tutor should not teach in university, then. :) Or, if the solution is correct (save for the value of the determinant), then maybe the statement of the problem is not a *faithful* translation of the German original? – 2017-02-14
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    The translation is correct. In lecture, we introduced it like this: Let $M \subset \Bbb R^n$ be a $k$-dimensional Manifold and $f: M \rightarrow \Bbb R$ a function. Now assume that there exists a bijective chart $\phi: T \rightarrow V \subset M \ (T \subset \Bbb R^k$ open) of $M$ such that $f|M \setminus V = 0$ and let $g$ be the gramian determinant with respect to $\phi$. Then, we want to call $f$ "integrable on $M$" if $t \rightarrow f(\phi(t) \sqrt g(t)$ is integrable on $T$. We define $\int_M f(x) d\sigma(x) := \int_T f(\phi(t)) \sqrt g(t) d^k t$. – 2017-02-14
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    It was then proven that this definition does not depend on the chart we choose. – 2017-02-14