Riemann integrable on the density set. $Q)$ The two real functions $f,g: I=[0,1] \to \mathbb{R} s.t.$ both$ f$ and $g$ are Riemann integrable on the $[0,1]$
Let $f(x) =g(x)$  for $\forall x \in A(\subset I)$ which is $\bar A = I(=[0,1])$ 
(Here the $\bar A$ is a closure of the $A$)
Show $\int_{0}^{1} f(x) dx = \int_{0}^{1}g(x)dx$ using the  refinement $P_n = \{ 0(=x_0), x_1, ...,x_n(=1) \}$

If we put $h=f-g$, then $h$ is Riemann integrable on the $I$.  So all I have to do show $\int_I h dx =0 $
What should I do next to show $\int_I h dx =0$ by $\vert U(h,P_n) - L(h,P_n) \vert < \epsilon $?
 A: As the function $h = f-g$ is Riemann integrable on $I = [0,1]$, for any $\epsilon > 0$ there exists a partition $P= \{x_0,x_1, \ldots, x_n: 0 = x_0 < x_1 < \ldots < x_n = 1\}$ such that
$$\tag{*}\left|\int_0^1 h(x) \, dx - S(P,h,\{t_k\})\right| < \epsilon,$$
where $S(P,h,\{t_k\}) = \sum_{k=1}^nh(t_k)(x_k - x_{k-1})$ is a Riemann sum with any choice of intermediate points $t_k \in [x_{k-1},x_k]$.  
Since $A$ is dense in $[0,1]$, i.e., $\bar{A} = [0,1]$ every subinterval $[x_{k-1},x_k]$ contains at least one point $t_k \in A$ where $h(t_k) = 0$ and, consequently, $S(P,h,\{t_k\}) = 0$.
Therefore, for every $\epsilon > 0$ we have
$$\left|\int_0^1 h(x) \, dx \right| < \epsilon,$$
which implies that $\displaystyle\int_0^1 h(x) \, dx = 0$.
A: Let $h = f-g$, as you said, also Riemann integrable. If $\int h \neq 0$, say w.l.o.g. $\int h >0$. Then there is a nondegenerate interval where $h>0$. (There is a step function that is lower than $h$, etc.) $f \neq g$ in the interval, so the closure of A cannot contain the interior of that interval, a contradiction.
