[ISI M.Math 2016, PMB, Group A] Suppose $f : [0,1] →\mathbb R$ is a bounded function such that $f$ is Riemann integrable on $[a,1]$ for every $a ∈ (0,1)$. Is $f$ Riemann integrable on $[0, 1]$ ? Justify your answer.
My attempt :
Consider a sequence ${{a_n}}_n$ from $[0,1]$ which converges to $0$. Since $0 \notin {{a_n}}_n$, $f$ is Riemann integrable in $[a_n,1]$ $\forall n \in \mathbb N$. Now, since $f$ is bounded in $[0,1]$, there is only one case of improper integration, i.e., having uncountable points on discontinuity in domain. Since $f$ is integrable in $[a_n,1]$, there is countable points of discontinuity. Let the set be $S_n$. Consider $\cup S_n = S$ and the set of all points of discontinuity of $f$ in $[0,1]$ is $B$. Then, $S \subseteq B$. Let a point $c \in B$ such that $c \notin S$. Since $c \in [0,1]$, we will find an $a_k$ (by Archimedean property) such that $a_k \leq c$. Then $c \in [a_k,1]$, which implies $S = B$, and hence the result.
I'm not convinced if there is any fault in my proofline. Any help would be appreciated.
There is no need to argue via points of discontinuity. One can proceed using Darboux sums.
Let $\epsilon >0$ be arbitrary. If $M>0$ is a bound for $|f|$ then split the interval $[0,1]$ into $[0,a]$ and $[a, 1]$ where $a=\min(\epsilon/4M,1)$. Since $f$ is Riemann integrable on $[a, 1]$ there is a partition $P'$ of $[a, 1]$ with $$U(P', f) - L(P', f)<\frac{\epsilon} {2} $$ Let $P=P'\cup \{0\}$ and then $P$ is a partition of $[0,1]$ and we have $$U(P, f) - L(P, f) < 2M\cdot\frac{\epsilon} {4M}+\frac{\epsilon} {2}=\epsilon $$ and therefore $f$ is Riemann integrable on $[0,1]$.