I want to prove that if $f$ is Riemann integrable on $[0,1]$ (i.e. $f\in\mathscr R$ on $[0,1]$) then it is true that $\int_0^1 f(x) \,dx=\lim_{c\to 0}\int_c^1 f(x)\, dx$.
It is known to me that since $f\in\mathscr R$ on $[0,1]$ then Riemann integrals $\int_c^1f(x)\,dx$ and $\int_0^c f(x)\,dx$ both exist for any $c\in (0,1)$, and we have:
$\int_0^1 f(x)\,dx=\int_0^c f(x)\,dx+\int_c^1f(x)\,dx$ ; and that there exist real numbers $M$ and $m$ such that $m\le f(x)\le M$ for all $x\in [0,1]$
Let's divide $[0,c]$ into $n$ equal parts to get the partition $P=\{0=t_0,t_1,\dots,t_{n-1},t_n=c\}$ such that $\Delta t_i=t_i-t_{i-1}=\frac cn$ for all $1\le i\le n$.
Let $M_i:=\sup\{f(t): x_{i-1}\le t\le x_i\}$ and $m_i:=\inf\{f(t): x_{i-1}\le t\le x_i\}$ for all $1\le i \le n$.
Since $\int_0^c f(x)\,dx$ exists, the following must hold for all partitions and in particular for $P$:
$L(P,f)\leq \int_0^cf(x)\,dx\le U(P,f)$
$\implies \sum_{i=1}^n m_i\Delta t_i\le \int_0^cf(x)\,dx\le \sum_{i=1}^n M_i\Delta t_i$
$\implies mc\leq \sum_{i=1}^n m_i\Delta t_i\le \int_0^cf(x)\,dx\le \sum_{i=1}^n M_i\Delta t_i\leq Mc$
whence it follows by Squeeze principal that $\lim_{c\to 0} \int_0^c f(x)\,dx=0$
The final result follows by limit rules.
Is my proof correct? Thanks.