To prove that the integral of the function lies between the lower darboux sum and the upper darboux sum m = inf(f(x): x is a member of [a,b]), M = sup(f(x): x is a member of [a,b])
suppose that the function is Riemann integrable 
Prove the following 
$$ m(b-a)\leq {\int_{a}^{b}} f(x) \, \mathrm{d}x \leq M(b-a) $$
Therefore what i have done was take the L(P,F) which results in m(b-a) and then i evaluated the U(P,f) which results in M(b-a), therefore since from the question the interval is closed b > a therefore this implies that since the value m is the inf which is the smallest value and the value M is the sup which is the largest value i can conclude that 
$$ m(b-a)\leq M(b-a) $$
which shows that the L(P,f) is less than or equal too U(P,f)
This is what  I have done thus far, now what i know is that for the Lower Riemann integral is taken as the sup (L(P,f) where p is a member of a Partition on [a,b]) therefore this value will increase and for opposite inf(U(P,f)) this value will decrease and the point at which both values are equal is known as the integral of the function. Therefore i can understand that the integral exists between the upper and lower sums once the integrals are equal but not sure how to show it can anyone assist.
I was considering using the riemann sum to show that is the limit exists and the function is said to be riemann integrable on the closed interval [a,b] then the limit of the riemann summ = the integral of the fucntion which lies between the upper and lower darboux sums.
Can anyone tell me if i am correct and if so am i on the right track to proving what is required?
 A: You are definitely on the right track.
Making this precise mathematically, we have for any partition $P = (x_0,x_1,\ldots, x_n)$ of $[a,b]$ and $\xi_j \in [x_{j-1},x_j]$ for all $j \in \{1,2\ldots,n\}$,
$$m = \inf_{x \in [a,b]}f(x) \leqslant \inf_{x \in [x_{j-1},x_j]}f(x)\leqslant f(\xi_j)  \leqslant \sup_{x \in [x_{j-1},x_j]}f(x) \leqslant  \sup_{x \in [a,b]}f(x) = M$$
Multiplying by $(x_j - x_{j-1})$ and summing leads to
$$\tag{*}m(b-a) \leqslant L(P,f) \leqslant S(P,f) \leqslant U(P,f) \leqslant M(b-a),$$
where $S(P,f)$ is a Riemann sum with (arbitrary) intermediate points $\xi_j$.
Given the hypothesis that $f$ is Riemann integrable, it must hold with $\|P\| := \max_{1\leqslant j \leqslant n}(x_j - x_{j-1})$ that
$$\tag{**}\lim_{\|P\| \to 0}S(P,f) = \int_a^bf(x) \, dx$$
Together (*) and (**) imply that
$$m(b-a) \leqslant \int_a^b f(x) \, dx \leqslant M(b-a)$$
More precisely 
For every $\epsilon > 0$ there exists $\delta > 0$ such that if $\|P\| < \delta$, then
$$\int_a^b f(x) \, dx - \epsilon < S(P,f) < \int_a^b f(x) \, dx+ \epsilon$$
Suppose that $M(b-a) < \int_a^b f(x) \, dx$.  Taking $\epsilon = \frac{1}{2}  \int_a^b f(x) \, dx +\frac{1}{2}M(b-a)$ we would have
$$S(P,f) > \int_a^b f(x) \, dx - \epsilon =  \frac{1}{2}  \int_a^b f(x) \, dx + \frac{1}{2}M(b-a) > M(b-a), $$
a contradiction.
By a similar argument we can show that $\int_a^b f(x) \, dx \geqslant m (b-a).$
