Does this proof of basic property of Riemann Integral hold? Hello I am wanting to know if this proof makes sense.
If $f$ is Riemann integrable on $[a,b]$ and $f$ is bounded on this interval by some $M \in \mathbb{R}$, show that $$\left\lvert \int_{a}^{b} f \right\rvert \le M(b-a).$$
I think it could also be proved using basic properties such as that if $f(x) \le g(x)$ then so, too, are the integrals, but I am especially wondering about the following.
By definitions we would have all $\epsilon \gt 0$ , $\exists \delta \gt 0$ such that for all partitions of mesh less than $\delta$
$$0 \le \left\lvert \sum_{i=1}^{n} f(t_{i})(x_{i}-x_{i-1})-L \right\rvert \lt \epsilon$$
but since $f$ is bounded,   
$$0 \le \left\lvert M\sum_{i=1}^{n}(x_{i}-x_{i-1})-L \right\rvert.$$ 
And
$$0 \le | M(b-a)-L |$$
since it must hold for any such partitions and tags.
Is this valid, does it make sense?
Thank you all
 A: You correctly argued that for sufficiently fine partitions we have
$$L - \epsilon \leqslant \sum_{k=1}^n f(t_k)(x_k - x_{k-1}) \leqslant L + \epsilon.$$
The ensuing steps, however, do not lead to the desired conclusion.  For example, $0 \leqslant |M(b-a) - L|$ does not guarantee that $|L| < M(b-a).$
To complete the proof, argue as follows.
If $L > M(b-a)$ choose $\epsilon$ such that $L - \epsilon > M(b-a)$ and you find sums greater than $M(b-a)$, a contradiction.  Make a similar argument to show we cannot have $L < -M(b-a)$.
Another approach is:
If $f$ is bounded, then $|f(x)| \leqslant M$ for $x \in [a,b],$ and
$$-M \leqslant f(x) \leqslant M.$$
Let $P = (x_0,x_1, \ldots, x_n)$ be any partition of $[a,b].$
Then for all $k = 1, 2, \ldots, n$ we have
$$-M \leqslant \inf_{x \in [x_{k-1},x_k]}f(x) \leqslant \sup_{x \in [x_{k-1},x_k]}f(x)\leqslant M.$$
Hence, for any lower and upper sum
$$-M(b-a) \leqslant L(P,f) \leqslant \int_a^b f(x) \, dx \leqslant U(P,f) \leqslant M(b-a),$$ and
$$\left|\int_a^b f(x) \, dx\right| \leqslant M(b-a). $$
A: My version in terms of the limit of sums interpretation of the Riemann integral:
Define $x_i=a+i/n$. As $f$ is bounded by $M$ on the interval, for all $n \in \mathbb{N}$
$$|\sum_{i=1}^{n}f(x^{*}_{i})[x_i-x_{i-1}]| \leq \sum_{i=1}^{n}|f(x^{*}_{i})[x_i-x_{i-1}]|\leq \sum_{i=1}^{n}|M[x_i-x_{i-1}]|$$
$$|\sum_{i=1}^{n}f(x^{*}_{i})[x_i-x_{i-1}]| \leq M(b-a) \,...(1)$$
Now as $f$ is Riemann integrable on $[a,b]$ then 
$$\lim_{n \to \infty} \sum_{i=1}^{n}f(x^{*}_{i})[x_i-x_{i-1}]=\int_{a}^{b}f(x)dx$$
So by $(1)$ then we have
$$|\int_{a}^{b}f(x)dx|\leq\lim_{n \to \infty}M(b-a)=M(b-a) $$
