# Direct Proof that a bounded, everywhere discontinuous function is not Riemann Integrable

I'm trying to construct a direct (that is, using properties of Riemann Integration and/or the partition definition of the Riemann Integral) prove that if a bounded function $$f: [a,b] \to \mathbb{R}$$ is discontinuous at every point on its domain, then it is not Riemann Integrable.

My initial approach has been to show that

$$\exists \epsilon >0$$ s.t. for all partitions $$P$$ of $$[a,b]$$,

$$U(f,P)-L(f,P) \geq \epsilon$$

I know this can be seen as a consequence that the interval $$[a,b]$$ has nonzero measure, so $$f$$ does not satisfy Lebesgue's criterion. However, how can I go about constructing such an epsilon for a general function f?

• Show instead that if $f:[a,b]\to\mathbb R$ is Riemann integrable there must exist at least one point of continuity. If Hankel could do it in 1870 you can do it in 2021. Use Riemann's criterion and a nested sequence of intervals. Dec 8, 2021 at 16:21
• @B.S.Thomson could you please clarify how to go about this? I don't understand how I could prove that the point in the intersection of those intervals would be continuous Dec 9, 2021 at 19:58
• Check this answer which proves that a Riemann integrable function must be continuous somewhere. Dec 11, 2021 at 3:48

Nobody jumping in to help so far ...

Here is a method you might like. At least it will give you some insight into continuity, if not the Riemann integral.

Preliminaries. There is some nice stuff to know about continuity. Let $$f:[a,b]\to\mathbb R$$ be an arbitrary function. Define

$$\phi(x,\delta) = \sup \{ |f(s)-f(t)|: s,t \in [a,b]\cap (x-\delta,x+\delta)\}$$ and

$$\phi(x) = \inf_{\delta>0} \phi(x,\delta).$$

Then

1. $$\phi(x)=0$$ if and only if $$f$$ is continuous at $$x$$.
2. Each set $$E_n= \{x\in [a,b]:\phi(x)\geq \frac1n\}$$ is closed.
3. The set of points of discontinuity of the function $$f$$ is the union $$\bigcup_{n=1}^\infty E_n$$ of this sequence of closed sets.
4. If $$f$$ is everywhere discontinuous one of these sets must contain a subinterval of $$[a,b]$$.

These are relatively elementary. For (1)-(3) see Section 6.7.2 The Set of Continuity Points in reference [1]. For (4) you need to know that no interval can be expressed as the union of a sequence of closed sets that are nowhere dense (i.e., none of them contain an interval). This is easy to prove using nested sequences of intervals. See Section 6.3 Nowhere Dense Sets in reference [1] if you wish to pursue this topic further (or any text that covers the Baire category theorem).

Now to prove our theorem:

Theorem. Let $$f:[a,b]\to\mathbb R$$ be discontinuous at each point. Then $$f$$ is not Riemann integrable.

Lemma. Suppose that $$f:[c,d]\to\mathbb R$$ has the property that $$\phi(x)\geq \delta>0$$ for every $$c. Then $$f$$ is not Riemann integrable on $$[c,d]$$.

This is particularly easy to prove from the elementary methods of the Riemann or Darboux integral. In fact your comments above about your "initial apporoach" will work here.

Proof of theorem. If $$f$$ is everywhere discontinuous then one of the sets $$E_n$$ defined above must contain an subinterval $$[c,d]$$ of $$[a,b]$$. By the lemma the function $$f$$ is not Riemann integrable on $$[c,d]$$ and hence cannot be integrable on $$[a,b]$$.

Other methods?

(a) Most students of mathematics are likely to respond this way: "Every Riemann integrable function must be continuous almost everywhere (as proved by Lebesgue) so of course this is true." See Section 8.6.3 Lebesgue’s Criterion of reference [1] for an accessible proof.

(b) Most 19th century mathematicians would have responded like this (since they had no notion of almost everywhere or measure zero sets): "If $$f$$ is Riemann integrable on $$[a,b]$$ then for every integer $$n$$ the set $$E_n= \{x\in [a,b]:\phi(x)\geq \frac1n\}$$ is a closed set of zero Peano-Jordan content. The set of discontinuities is the union of this sequence and that cannot be all of $$[a,b]$$."

(In fact Lebesgue's criterion is a nearly trivial extension of the 19th century version.)

(c) A direct proof (already suggested) is to use Riemann's criterion and a nested sequence of intervals to uncover one point of continuity. The user Paramandand Singh supplied this link for those details: $f \in {\mathscr R[a,b]} \implies f$ has infinitely many points of continuity.

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