Show that $f\varphi_A$ is contained in $D \cup \partial A$ Hey guys i'm having a bit of trouble with this one. I'm hoping that you guys can help me out. 
If $D$ is the set of discontinuities of $f:R^n \rightarrow R$, show that the set of discontinuities of $f\varphi_A$ is contained in $D \cup \partial A$
Where $\varphi_A$ is the characteristic function of $A$
thanks as always guys
 A: I'll recall some definitions, give some motivation for the exercise and give a hint.
Well, first of all, recall why do we introduce that characteristic function: the integral is easy to define on rectangles, however we want to extend to more sets. So, what we do is construct a function $\varphi_A : \mathbb{R}^n \to \mathbb{R}$ defined by $\varphi_A(x)=1$ if $x \in A$ and $\varphi_A(x)=0$ otherwise. This function indicates if $x$ is or not in $A$.
Now consider a function $f: U \subset \mathbb{R}^n \to \mathbb{R}$ where $U$ is arbitrary. The function $f\varphi_U$ is a function $(f\varphi_U)(x)=f(x)$ for points inside $U$ and and $(f\varphi_U)(x)=0$ for points outside $U$. We are interested in the following: given such $f$, we want to consider some rectangle $A$ such that $U \subset A$ and such that we can extend $f$ to $A$ by this construction with the characteristic function, in other words: the extended function will accord with $f$ on $U$ and be zero on the rest of the rectangle.
We then define the integral over a general set $U$ to be:
$$\int_Uf=\int_A f\varphi_U$$ 
The RHS is the integral over a rectangle, so this expression will have sense if $f\varphi_U$ is integrable on $A$. For that we need that the set of discontinuities of $f\varphi_U$ has measure $0$. I think this gives you good understanding on what the exercise wants. I think that a good place to start is to find out the set of discontinuities of $\varphi_U$, can you find out where this function is discontinuous?
