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



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?

  • $\begingroup$ can you explain more $\endgroup$ – user229886 Feb 11 '17 at 7:56
  • $\begingroup$ Added a question. Can you answer that. Do you know the Lebesgue criterion for Riemann integrability? $\endgroup$ – RRL Feb 11 '17 at 7:58
  • $\begingroup$ 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? $\endgroup$ – user229886 Feb 11 '17 at 8:01
  • $\begingroup$ Since $f$ is integrable on $[1/n,1]$, the measure of $D_f([1/n,1])$ is $0$. Also $m (U_n A_n) \leqslant \sum_n m(A_n)$. $\endgroup$ – RRL Feb 11 '17 at 8:03
  • $\begingroup$ Isn't this leading up to the fact that $f$ must be discontinuous only on a set of measure zero, at worst? $\endgroup$ – RRL Feb 11 '17 at 8:08

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 $|f(x) |<M$ for all $x\in[0,1]$. Now $f$ is Riemann integrable on interval $[\epsilon /4M,1]$ and hence there is a partition $P'$ of $[\epsilon /4M,1]$ such that $$U(f, P') - L(f, P') <\frac{\epsilon} {2}\tag{1}$$ Now consider $P=\{0 \} \cup P'$ so that $P$ is a partition of $[0,1]$ and $$U(f, P) - L(f, P) = (A-B) \frac{\epsilon} {4M}+U(f,P')-L(f,P')\tag{2}$$ where $$A=\sup\, \{f(x) :x\in[0,\epsilon /4M]\},\,B=\inf\,\{f(x):x\in[0,\epsilon /4M]\}$$ Clearly $A-B\leq 2M$ and hence it follows from equations $(1)$ and $(2)$ that $$U(f, P) - L(f, P) <\epsilon $$ and thus our job is done.

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

  • 1
    $\begingroup$ I think this is pedagogically a nice answer. Upvoted. $\endgroup$ – RRL Feb 11 '17 at 19:10

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