Show that $f$ is integrable and $\int_{I} f=0$ Let $I$ b a generalized rectangle in $\Bbb R^n$ 
Suppose the bounded function $f:I\to \Bbb R$ assumes the value $0$ except at a single point $x \in I$ 
Show that $f$ is integrable and $\int_{I} f=0$ 

I think that 
Let $\epsilon >0$ 
Let's choose a partition $$\vert \vert P\vert\vert ^n< \frac{\epsilon}{2\vert f(x)\vert} $$
and x contained in exactly one generalized rectangle in $P$. 
I need to calculate $L(f,P)$ and $U(f,P)$. 
How to calculate $L$ and $U$? Please help me for the evaluation. Thank you. 

Assume that $f\geq 0$; if not, since it is only non-zero at $x$, you can replace $f$ by $-f$.
Then every (generalized) rectangle in your partition that does not contain $x$ has both upper and lower sum 0. For the rectangle that contains $x$, note that all $y$ in the rectangle satisfy 
$$
0\leq f(y)\leq f(x),
$$
and so $x$ must actually be maximum on the rectangle. Hence the lower sum is 0, and the upper sum is $f(x)\cdot V_P$, where $V_P$ is the volume of the box containing $x$.
Hence for any partition $P$ in which $x$ lies in only one box, we have
$$
L(f,P)=0\qquad\text{and}\qquad U(f,P)=f(x)\cdot V_{P}.
$$
