Proving a constant function $f(x) = c$ is Riemann integrable Prove that a constant function $f(x) = c$, where $c$ is in the Real Numbers, is Riemann integrable on any interval $[a, b]$ and $\int_a^bf(x) dx = c(b-a)$.

By looking at the definition, it looks like I am going to explain that it's bounded (which would be obvious since the function is constant?)  Additionally, it would appear that $\inf(f)$ and $\sup(f)$ also obviously exist since $f$ is constant.  The parts I am having trouble understanding involve explaining that $\sup\{L(P,f)\} = \inf\{U(P,f)\}$, as well as proving that $\int_a^bf(x) dx = c(b-a)$.  Let me know if what I have so far is okay, and please give me some guidance on the rest.  Thanks! 
 A: You're on the right track so far. For the sup=inf part, you're in a really nice position: if you take any partition and compute the max of f on each subinterval, you'll get c. So $U(P, f) = \sum_i c \cdot (t_{i+1} - t_i) = c \cdot \sum_i  t_{i+1} - t_i = c(b-a)$. That means that EVERY SINGLE Upper sum turns out to be $c(b-a)$. I'll bet that you can compute the inf of a set that contains only a single number, right? Then you're on your way.  
A: Sorry, this answer is coming in after five years.
Consider, for each $n$, a uniform partition, $P_n$ on $[a,b]$ which is a finite sequence of real numbers such that 
$$a=x_0<x_<\cdots<x_n=b.$$ 
Define $I_j=[x_{j-1},x_j)$ for each $1\leq j\leq n-1\,$ and $\,I_n=[x_{n-1},x_n]$. Clearly, $f$ is bounded on $[a,b]$, so $m:=\inf f$ and $M:=\sup f$ both exist. Fix $j\in \{1,2,\cdots,n\}$  and $\,x\in I_j,$ then 
$$m_j\leq f(t_j)\leq M_j,\;\;\text{where}\;\;t_j\in I_j,$$
\begin{align}m_j=\inf\{f(x):\;x\in I_j  \}=c \;\;\text{and}\;\; M_j=\sup\{f(x):\;x\in I_j  \}=c.\end{align}
Summing up to $n,$ results to
 \begin{align}L(f,P_n)&=c(b-a)=c(x_{n}-x_{0})=c\sum^{n}_{j=1}(x_{j}-x_{j-1})=\sum^{n}_{j=1}m_j(x_{j}-x_{j-1})\leq \sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})\\&\leq \sum^{n}_{j=1}M_j(x_{j}-x_{j-1})=c\sum^{n}_{j=1}(x_{j}-x_{j-1})=c(x_{n}-x_{0})=c(b-a)=U(f,P_n),\end{align} 
Thus,
\begin{align}\sup\limits_{P_n} L(f,P_n)&=\lim\limits_{n\to \infty}L(f,P_n)=c(b-a)\leq \lim\limits_{n\to \infty} \sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})\\&\leq c(b-a)=\lim\limits_{n\to \infty}U(f,P_n)=\inf\limits_{P_n} U(f,P_n),\end{align}
and
\begin{align}\int^{a}_{b}f(x)dx= \lim\limits_{n\to \infty}\sum^{n}_{j=1}f(t_j)(x_{j}-x_{j-1})=\sup\limits_{P_n} L(f,P_n)=\inf\limits_{P_n} U(f,P_n)=c(b-a),\end{align} 
as required.
A: (By using Darboux Sums)
Let $\mathcal{P}=\{x_0,x_1,\ldots,x_n\}$ be a partition of $[a,b]$.
If $m_k$ and $M_k$ denote the infimum and supremum of $f$ on $[x_{k-1},x_k]$, respectively, then $m_i=M_i=c$ for all $k=1,2,\ldots,n$.
Thus
\begin{equation}
L(f;\mathcal{P})=\sum_{i=1}^n m_i (x_i-x_{i-1})=c(b-a).
\end{equation}
Since this is true for any $\mathcal{P}\in \mathscr{P}[a,b]$, taking the supremum of the lower sum over all partitions of $[a,b]$ gives
\begin{equation}
L(f):=\sup_{\mathcal{P}\in \mathscr{P}[a,b]}L(f;\mathcal{P})≥c(b-a)
\end{equation}
Similarly,
\begin{equation}
U(f)≤c(b-a),
\end{equation}
from whence it follows 
\begin{equation}
c(b-a)≤L(f)≤U(f)≤c(b-a).
\end{equation}
Therefore, $L(f)=U(f)=c(b-a)$, and this proves the assertion. $\blacksquare$
A: Constant function is a continoues function and we know that every continoues function is a Riemann integrable.Since Constant function is a Riemann Integrable.
A: (By using Riemann Sums)
Let $\dot{\mathcal{P}}=\{([x_{i-1},x_i],t_i)\}_{i=1}^n$ ̇be any tagged partition of $[a,b]$. Then
\begin{equation}
S(f;\dot{\mathcal{P}})=\sum_{i=1}^nf(t_i )(x_i-x_{i-1})=c\sum_{i=1}^n(x_i-x_{i-1})=c(b-a).
\end{equation}
Hence, for any $\epsilon>0$, we can choose $\delta_\epsilon=1$ so that for any tagged partition $||\dot{\mathcal{P}}||<\delta_\epsilon$, then
\begin{equation}
|S(f;\dot{\mathcal{P}})-c(b-a)|=0<\epsilon.
\end{equation}
Since $\epsilon>0$ is arbitrary, we conclude that $f\in \mathcal{R}[a,b]$ with 
\begin{equation}
\int_a^b f=c(b-a). \blacksquare
\end{equation}
