Riemann integrability of $f$ on a [0,1]

Question

Suppose $f:[0,1] \rightarrow \mathbb{R}$ is a bounded function such that $f$ is Riemann integrable on $[a,1]$ for every $a \in (0,1)$. Is $f$ Riemann integrable on [0,1]?

Answer 1

Proposition: Let $f:[a,c]\rightarrow \mathbb{R}$ be a bounded function.

Suppose that $f \in R[a,b], \forall a<b<c,$ so $f \in R[a,c]$

Proof:

Let $\epsilon>0$ and define $b=c-\frac{\epsilon}{2\omega(f,[a,c])}.$

$f \in R[a,b] \Rightarrow$ there exists partition $\Pi_1$ of the interval $[a,b]$ such that $\omega(f_{[a,b]},\Pi_1)<\frac{\epsilon}{2}$.

Let us look at the partition: $\Pi=\Pi_1 \cup\{c\} $ of the interval $[a,c]$.

We have:

$$\omega(f,\Pi)=\omega(f_{[a,b]},\Pi_1)+\frac{\epsilon}{2\omega(f,[a,c])}\omega(f,[b,c])\leq\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon$$

By Darboux, we get $f\in R[a,c] \ _\blacksquare$

Answer 2

Yes. We may look at it from a measure-theoretic viewpoint (no need to be scared by "measure theory", which is in fact more natural and simple). Henri Lebesgue has got a theorem asserting that a bounded function is Riemann integrable iff it is continuous almost everywhere. By assumption $f$ is continuous almost everywhere on $[a,1]$ for all $0 < a < 1$. Let $0 < a_{k} < 1$ for all $k \in \mathbb{N}$; let $a_{k} \to 0$; let $D_{k}$ be the set of discontinuities of $f$ over $[a_{k}, 1]$ for all $k \in \mathbb{N}$; let $D$ be the set of discontinuities of $f$ over $[0,1]$. Then $D = \bigcup_{k}D_{k} \cup \{ 0 \}\ \text{or}\ = \bigcup_{k}D_{k}$; so $D$ has Lebesgue measure $0$ too. So by the Lebesgue's theorem the function $f$ is Riemann integrable over $[0,1]$.