Proving a bounded function is integrable on $[a,b]$ I need to prove that

Let $f:[a,b] \to \mathbf{R}$ be a bounded map and let $f$ be an integrable map on the interval $[c,b]$ for all $c \in (a,b).$ Then, $f$ is integrable on $[a,b].$

My attempt:
By hypotesis, $f:[a,b] \to \mathbf{R}$ is bounded, so there exists $0 < K \in \mathbf{R}$ such that $|f(x)| \leq K$, for all $x \in [a,b].$ Let $\varepsilon>0,$ and take $c \in (a,b)$ such that $K \cdot (c-a) \leq \frac{\varepsilon}{4}$. By hypotesis, $f$ is integrable on $[c,b],$ then, there exists a partition $\left\{t_1, \dots, t_n \right\} \subset [a,b]$ such that $\sum_{i=2}^{n} \omega_i(t_i-t_{i-1}) < \frac{\varepsilon}{2}$. Let's make $t_{1}=a$, then we get  a partition $\left\{t_0, t_1, \dots, t_n \right\}$ of $[a,b]$.
Now I need to prove that $\omega_0 \leq 2K$, for me to get that $f$ is integrable on $[a,b]$. Am I following the correct idea?
I don't know how to procceed. Any ideas would be appreciated!
 A: I find it easier to talk about the equivalent in Darboux integral terms, that a funct ion is integrable if the upper sum minus the lower sum for any partition is less than any $\epsilon$ you want.  So to show intgrability over $[a,b]$ we start with $\epsilon>0$.  The trick is to find the right $c$ to use so that when we integrate from $[a,c]$ we are guaranteed to be less than $\frac \epsilon 2$,  since we can make the part from $[c,b]$ be less than $\frac \epsilon 2$ by the fact that it is integrable.  That's easy,  just figure out worst case error, if $M>0$ is the bound on $|f(x)|$, then the biggest error change from upper to lower is $2M$ over a width of $c-a$, so the integral on $[a,c]$ is bounded by $2M(c-a)$.  I find it easier to focus on the width, so define $\delta=c-a$
Setting that to $\frac \epsilon 2$ and solving for $\delta$ we get
$$2M\delta<\frac \epsilon 2$$
$$\delta<\frac \epsilon {4M}$$
Now you just have to make sure that $a+\delta<b$,  so use $\delta'=\min \{\delta,b-a\}$
now we can safely partition our integral on $[a,b]$ that on $[a,a+\delta']\cup [a+\delta',b]$ and be guaranteed that both parts are below $\frac \epsilon 2$ in the difference between upper and lower sums
A: So I guess I've found out:
By hypotesis, $f:[a,b] \to \mathbf{R}$ is bounded, so there exists $0 < K \in \mathbf{R}$ such that $|f(x)| \leq K$, for all $x \in [a,b].$ Let $\varepsilon>0,$ and take $c \in (a,b)$ such that $K \cdot (c-a) \leq \frac{\varepsilon}{4}$. By hypotesis, $f$ is integrable on $[c,b],$ then, there exists a partition $P=\left\{t_1, \dots, t_n \right\} \subset [a,b]$ such that $\sum_{i=2}^{n} \omega_i(t_i-t_{i-1}) < \frac{\varepsilon}{2}$. Let's make $t_{n+1}=b$, then we get  a partition $P'=\left\{t_0, t_1, \dots, t_n, t_{n+1} \right\}$ of $[a,b]$, with $\omega_{n+1}$.
Note that $\omega_{n+1}$ is bounded once $f$ is bounded on $[c,b]$, by hypotesis. Then we get $\omega_{n+1}<2K$. Therefore, $\omega_{n+1} \cdot (t_{n+1}-t_n)=\omega_{n+2}(b-c)<\frac{\varepsilon}{2}$.
Now we conclude that $\sum_{i=2}^{n} \omega_i(t_i-t_{i-1}) < \varepsilon$ then $f$ is integrable on $[a,b]$.
