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The problem is as follows (from a multivariable calculus course, so preferably a solution without deep measure-theory):

Let $f: \mathbb{R} \rightarrow \mathbb{R}$ be a continuously differentiable function. We define $$\phi:\mathbb{R}^2 \rightarrow \mathbb{R}^2, \phi(x, y)=(x+f(x+y), y - f(x+y)).$$

Prove that the Jordan measure (area) of any Jordan measurable set $E$ is equal to $\phi(E)$.

My experience with the Jordan measure is only integral-related, so I'm unfamiliar with any of its properties beyond the fact that it can be used to define areas and volumes. Also, I seem to remember that it can be defined analogously to the Lebesgue measure as follows:

$$\mu(E)=\inf\{\sum_{k=1}^{n}\mu(B_{k}) | B_{k} \space \textrm{is a finite sequence of open balls such that} \space E \subseteq \bigcup\limits_{k=1}^{n} B_{k}\},$$ but I'm neither sure that this is useful, nor 100% sure if it's true, for that matter. Any hint would be appreciated.

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This is a preliminary answer and I'm grateful for any (even the most minute) criticisms, nitpickings and comments on my solution, as I barely know what Jordan measures are (although I believe I got the gist of it).

If $E$ is Jordan measurable, then the integral $$I=\iint_E \,dx\,dy$$ exists. By substituting $(x,y) \mapsto (x+f(x+y),y-f(x+y))$, we get $$I=\iint_{\phi(E)}J(\phi)dxdy,$$ where $J(\phi)$ (extremely informaly) denotes the Jacobian of the substitution, which is equal to: $$\begin{vmatrix} 1+f'(x+y) & f'(x+y)\\ -f'(x+y) & 1-f'(x+y) \end{vmatrix}=1,$$ which completes the proof.

Are there any auxiliary statements that need to be proven or at least noted before I can claim that this substitution will work?

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    You are using the cov theorem. Proving that the transformation is bijective is required (but easy). And $F = \phi$. – 2017-03-03
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    I googled cov theorem and nothing came up except covariance. Would you mind elaborating on what exactly that theorem's called, if it has a full name? Also, which transformation do you mean? $\phi$? – 2017-03-03
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    Yes, the transformation is $\phi$. See https://en.wikipedia.org/wiki/Integration_by_substitution#Substitution_for_multiple_variables. – 2017-03-03