Integrability question with a function on a box in $\mathbb{R}^2$ (bounty added) Let $f:\mathbb{R} \to\mathbb{R} \ be \ bounded,$ 
$\phi: \mathbb{R}^2 \to\mathbb{R}^2 $ be defined as $\phi(x,y)=(x,y+f(x))$

Prove that if for every bounded box $B\subset \mathbb{R}^2, \phi(B)$
  is admissible (that means, $\mathbb{1}_{\phi(B)}$ is
  Riemann-integrable), f is continuous almost everywhere.

My work:
I tried to assume f is not continuous almost everywhere.
That means that there is a segment in $\mathbb{R}$ $[x1,x2]$ where f is not riemann-integrable. so we look at that segment and take the box defined by $(x1,y1),(x2,y2),(x2,y1),(x1,y2)$. We activate $\phi$ on that box.
Now here is the part I'm not sure of..
Because $f$ is bounded (and that f is not riemann integrable in this segment) we can take a $y_c$ so that $y_c$ is between $y1+f(x)$ and $y2+f(x)$ in a dense set of the segment $[x1,x2]$ and is not between those values in a different dense set in that segment.
Now we see that the Upper darboux sum in that segment of $\mathbb{1}_{\phi(B)}(x,y_c)=1*(x2-x1)$
and the Lower darboux sum is 0.
That would mean we found a box B where $\phi(B)$ is not admissible.
And that proves it.

This is what I answered in my test and I lost 30 out of 40 points on
  it, so I would appreciate it a lot if you tell me where I was wrong
  and why.
Thanks a lot.

 A: This is how I would tackle this, maybe you can draw some conclusions from that.
Let $S$ be the set of discontinuities of $f$ and assume $\mu(S)>0$.
For $\epsilon>0$ let $S_\epsilon$ be the set of points $x$ such that there are sequences $\xi_i\to x$ and $\eta_i\to x$ such that $\lim f(\xi_i)$ and $\lim f(\eta_i)$ exist and differ by more than $\epsilon$. Then $S=\bigcup_{n\in\Bbb N} S_{1/n}$ (this uses that $f$ is bounded).
Hence at least one $S_{1/n}$ has positive measure.
Also, for suitable $x_1<x_2$, the set $S_{1/n}\cap (x_1,x_2)$ has positive measure $c$. Let $B=(x_1,x_2)\times(0,\frac1{3n})$.
If we partition a box containing $\phi(B)$ with a rectangular $u$ by $v$ grid (with $v<\frac1{3n}$), then in each of the at least $c/u$ column interiorly affected by $S_{1/n}\cap (x_1,x_2)$, we find at least $\frac2{3nv}$ grid rectangles that contain both points $\in \phi(B)$ and point $\notin \phi(B)$. Hence the Riemann upper and lower sum for $1_{\phi(B)}$ for this particular partition differ by at least $\frac{2c}{3n}$. As this does not depend on $u,v$, we conclude that $1_{\phi(B)}$ is not Riemann integrable.
