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Suppose $f:[0,1]\rightarrow\mathbb{R} $ be a bounded function such that, $f:[a,1]\rightarrow\mathbb{R} $, for all $a \in (0,1)$, is Riemann Integrable .Then what can you say about Riemann Intagrability of $f:[0,1]\rightarrow\mathbb{R} $..

What I feels is it is not Riemann integrable because we can construct such a function which is not Riemann integrable at 0, like Dirichlet function around Zero, but i cant proceed with it

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Hint:

Let $D_f(A)$ denote the set of discontinuity points in $A \subset [0,1]$.

Then we have

$$D_f((0,1]) = \bigcup_n D_f([1/n,1])$$

What can be said about $D_f([1/n,1])$ and this countable union?

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    can you explain more – 2017-02-11
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    Added a question. Can you answer that. Do you know the Lebesgue criterion for Riemann integrability? – 2017-02-11
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    But How can you say about non-zeroness of measure of D. Then only one can say about that function is not Riemann integrable Right? – 2017-02-11
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    Since $f$ is integrable on $[1/n,1]$, the measure of $D_f([1/n,1])$ is $0$. Also $m (U_n A_n) \leqslant \sum_n m(A_n)$. – 2017-02-11
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    Isn't this leading up to the fact that $f$ must be discontinuous only on a set of measure zero, at worst? – 2017-02-11
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    I Got that But can you five me an example – 2017-02-11
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    I don't know what example you are looking for. The set of discontinuities lie in a union of measure zero sets. Therefore $D_f((0,1])$ has measure zero -- and it is still measure zero if we add the point $\{0\}$. So what does Lebesgue now tell us about the integral of $f$ over $[0,1]$? – 2017-02-11
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    OOhk... Fine..!! I finally Got you – 2017-02-11
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    @Master X: Excellent! – 2017-02-11
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    I gave another approach which avoids Lebesgue's criterion and instead uses the upper and lower Darboux sums. +1 – 2017-02-11
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It is not difficult to prove directly that under given conditions $f$ is Riemann integrable on $[0,1]$. To prove this we will show that corresponding to any given $\epsilon>0$ there is a partition $P$ of $[0,1]$ such that the difference between upper and lower Darboux sum for $f$ over $P$ is less than $\epsilon $.

Since $f$ is bounded on $[0,1]$, there is a number $M$ such that $|f(x) |

Nitpick: Someone may ask "What happens when $\epsilon/4M\geq 1$?" Then we can see that the partition $P=\{0,1\}$ works.

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    I think this is pedagogically a nice answer. Upvoted. – 2017-02-11