Doubt in finding the integral of $f(x)=x$ using the upper and lower sum definition of the integrals. The definiton of integrals read 

A function $f$ which is bounded on $[a,b]$ is integral-able on $[a,b]$ if and only if for every $\epsilon$ there exist a partition P such that $$U(f,P) - L(f,P) \lt \epsilon$$ and the integral is the number, and it is unique, that lies between all lower and upper sums. 

($L(f,P)$ meant the lower sum of $f$ on the partition P, and similarly the $U(f,P)$) 
So, we want to find the integral of function $f(x) =x$ from $[0,b]$. Let's divide the interval $[0,b]$ into $n$ equal subintervals, that is $$P = \{0=t_0, t_1 , ...., t_n=b\} \\
\textrm{such that}
\\ t_0 = 0\\ t_1= \frac{b}{n}\\ ... t_i = \frac{ib}{n} $$ 
in any sub-interval $\{t_{i-1}, t_i\}$ the minimum of $f$, $m_i$, will be $f(t_{i-1})= t_{i-1}$ and maximum of $f$, M_i, will be $f(t_i) = t_i$ (because $f$ is monotonously increasing). Now, defining the upper and lower sums 
$$
L(f,P) = \sum_{i=1}^{n} m_i (t_{i} - t_{i-1}) 
L(f,P ) = \sum_{i=1}^{n} t_{I-1} (\frac{b}{n}) = \frac{b^2}{n^2} \sum_{i=1}^{n} (i-1)\\
L(f,P) = \frac{b^2}{n^2} \frac{(n-1)n}{2}$$ 
$$U(f,P) = \sum_{i=1}^{n} M_i (t_i - t_{i-1}) \\
U(f,P) = \frac{b^2}{n^2} \frac{n(n+1)}{2}$$ 
Writing them more specifically we have $$ L(f,P) = \frac{b^2}{2n} (n-1) \\ U(f,P) = \frac{b^2}{2n} (n+1)$$ 
$$U(f,P) - L(f,P) = \frac{b^2}{n}$$ Now, I have got the difference between upper and lower sum, and it is equal to $\frac{b^2}{n}$ which can be made as small as we wish, so our function is integral-able.   
Now, how to find this integral? The book writes 

It is clear, first of all, that $$L(f,P_n) \leq \frac{b^2}{2} \leq U(f,P_n) ~~~~~~~~~~~~~~~~~~\text{for all $n$}$$
  This inequality shows only that $b^2/2$ lies between certain special upper and lower sums, but we have just seen that $U(f,P)-L(f,P)$ can be made as small as desired , so there is only one number with this property. Since the integral certainly has this property, we can conclude that $$\int_{0}^{b}= \frac{b^2}{2}$$. 

I really couldn't understand how $\frac{b^2}{2}$ will be the only number in between. I'm having hard time in understanding what book is trying to say. Can someone please explain the workings of book or his/her own method? 
 A: If I understand what you are given, you have
$$
L(f,P_n)\le\frac{b^2}2\le U(f,P_n)
$$
and
$$
U(f,P_n)-L(f,P_n)\le\frac{b^2}n
$$
That means that
$$
\begin{align}
L(f,P_n)
&\ge U(f,P_n)-\frac{b^2}n\\
&\ge\frac{b^2}2-\frac{b^2}n
\end{align}
$$
and that
$$
\begin{align}
U(f,P_n)
&\le L(f,P_n)+\frac{b^2}n\\
&\le\frac{b^2}2+\frac{b^2}n
\end{align}
$$
Therefore,
$$
\frac{b^2}2-\frac{b^2}n\le L(f,P_n)\le U(f,P_n)\le\frac{b^2}2+\frac{b^2}n
$$
and we can apply the Squeeze Theorem.
A: Here is the important theorem which may seem obvious but is difficult to prove:

Theorem: If a function $f:[a, b] \to\mathbb{R} $ is bounded on $[a, b] $ then $$\lim_{|P|\to 0}U(f,P)=\inf\,\{U(f,P)\mid P\text{ is a partition of }[a, b] \} $$ and $$\lim_{|P|\to 0}L(f,P)=\sup\,\{L(f,P)\mid P\text{ is a partition of }[a, b] \} $$

Using this theorem the job of finding the supremum of lower sums (or infimum of upper sums) is reduced to finding the limit of corresponding sums as norm of partition tends to $0$. In particular we can take a sequence of partitions which are uniform (all subintervals of equal length) such that number of subintervals tends to $\infty $.
Thus for your example it is sufficient to take the limit of $U(f, P_n) $ and $L(f, P_n) $ as $n\to\infty $. If these limits are equal (as is the case here) the function is Riemann integrable with the common limit ($b^2/2$) as its integral.

However the argument given in your question is almost correct (may be it needs a little more effort) and you shouldn't doubt yourself (not everything in analysis is difficult and even if they are you can't be wrong every time).
Here is a way to add some detail to make it perfect (from almost correct). Start with the observation that every lower sum is less than or equal to any specific upper sum. Thus for any partition $P $ we must have $$L(f, P) \leq U(f, P_n) \tag{1}$$ for all $n$ and letting $n\to\infty $ we have $$L(f, P) \leq \frac{b^2}{2}\tag{2}$$ Why does taking limit as $n\to\infty $ work for us?? Well you should further notice that as $n$ increases $U(f, P_n) $ decreases strictly and thus infimum of all $U(f, P_n) $ equals its limit $b^2/2$. And if it were possible that $L(f, P) > b^2/2$ then by definition of infimum we would have some value of $n$ such that $$b^2/2\leq U(f, P_n) < L(f, P) $$ contradicting that any lower sum can't exceed any upper sum. And thus we must have $L(f, P) \leq b^2/2$ and this justifies the process of taking limits as $n\to\infty $ and deduce equation $(2)$ from $(1)$.
In a similar manner $$U(f, P) \geq L(f, P_n) $$ gives us $$U(f, P) \geq\frac{b^2}{2}$$ Combining all the above this proves that $$L(f, P) \leq \frac{b^2}{2}\leq U(f, P') \tag{3}$$ for any partitions $P, P'$. Notice that from the result $$L(f, P_n) \leq \frac{b^2}{2}\leq U(f, P_n)$$ established in your question we can infer the stronger result $(3)$ above.
Since you have proved that the function is Riemann integrable not more than a single number can lie between all lower and all upper sums (this is also mentioned in your question which means you understand this part). What have we got now on our hands? Well, just that the integral value is $b^2/2$. Done!!

The argument presented above (in second part) can be abstracted out by stripping all details of Riemann integral and partitions to lead to a simple

Lemma: Let $A, B$ be non-empty subsets of $\mathbb{R} $ such that no member of $A$ exceeds any member of $B$. If there exist sets $C, D$ such that $$C\subseteq A, D\subseteq B, \sup C=\inf D$$ then $\sup A=\inf B$.

The sets $C, D$ can also be replaced by some sequences $x_n\in A, y_n\in B$ with $\lim x_n=\lim y_n$. 
And now this looks so simple/trivial/obvious and you may easily prove it. 
