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]?
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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 $[0,1]$.
Yes
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1Actually, if $0$ is a discontinuity point, then $D = \bigcup_{k}D_{k} \bigcup {0}$. Of course, it doesn't alter the correctness of your proof, since {0} has 0 Lebesque measure. – Apr 09 '17 at 10:15
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@EugenCovaci, Yes thank you. It is my typo. But pointing out this is for other readers good :). – Yes Apr 09 '17 at 10:33
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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 $f\in R[a,c] \ _\blacksquare$
