Ngô Quốc Anh

April 5, 2015

The set of continuous points of Riemann integrable functions is dense

Filed under: Uncategorized — Ngô Quốc Anh @ 15:03

In this note, we prove that the set of continuous point of Riemann integrable functions f on some interval [a,b] is dense in [a,b]. Our proof start with the following simple observation.

Lemma: Assume that P=\{t_0=a,...,t_n=b\} is a partition of [a,b] such that

\displaystyle U(f,P)-L(f,P)<\frac{b-a}m

for some m; then there exists some index i such that M_i-m_i < \frac 1m where M_i and m_i are the supremum and infimum of f over the subinterval [t_{i-1},t_i].

We now prove this result.

Proof of Lemma: By contradiction, we would have M_i-m_i \geqslant \frac 1m for all i; hence

\displaystyle \frac{b-a}m = \sum_{i} \frac{t_i-t_{i-1}}{m}\leqslant \sum_{i} \big(M_i-m_i\big)(t_i-t_{i-1})=U(f,P)-L(f,P),

which gives us a contradiction.

We now state our main result:

Theorem. Let f be Riemann integrable over [a,b]. Define

\displaystyle \Gamma = \{ x\in [a,b] : f \text{ is continuous at } x\}

Then \Gamma is dense in [a,b].

We now prove the theorem.

Proof of Theorem: Let P be a partition of [a,b] such that U-L<b-a, where U and L are the upper and lower Darboux sums of f. By the lemma, there is some index i for which M_i - m_i<1.

Step 1. Narrow [a,b] to obtain a subinterval whereas \sup f - \inf f is relatively small.

To see this, we first select

\displaystyle I_1=(a_1,b_1)=\begin{cases}(t_{i-1}, t_i), & \text{ if } i \ne 1 \text{ or } i\ne n\\ (a_1, t_1), & \text{ if } i =1 \text{ and for some } a_1 \in (a,t_1) \\ (t_{n-1}, b_1), & \text{ if } i =n \text{ and for some } b_1 \in (t_{n-1}, b). \end{cases}

If I_1 happens to be (t_{i-1}, t_i), then we immediately get \sup_{I_1} f -\inf_{I_1}f <1. Otherwise, it will be the case that

\displaystyle \sup_{A'} f \geqslant \sup_A f

where A'=[t_0,t_1] (resp. [t_{n-1},t_n]) and A=[a_1,t_1] (resp. [t_{n-1},b_1]) and also that

\displaystyle\inf_{B'} f \leqslant \inf_B f

where B'=[t_0,t_1] (resp. [t_{n-1},t_n]) and B=[a_1,t_1] (resp. [t_{n-1},b_1]). Thus, in any case, there would exist a<a_1<b_1<b and

\displaystyle \sup_{I_1} f -\inf_{I_1}f <1

where I_1=[a_1,b_1]. Since f is integrable over [a,b], it is integrable over [a_1,b_1].

Step 2. Repeat the above argument to get a sequence of strictly nested subintervals.

The very same reasoning but with m=2 gives us a_2, b_2 with a<a_1<a_2<b_2<b_1<b and with

\displaystyle \sup_{I_2} f-\inf_{I_2}f< \frac 12

Continuing the process gives us a sequence of nested closed intervals \{I_n:n\in\mathbb N\} such that

\displaystyle \sup_{I_n}f - \inf_{I_n} f< \frac 1n.

By the Cantor theorem,

\displaystyle \bigcap_{n \in \mathbb N} I_n \ne \emptyset.

Step 3. Existence of points of continuity in every subinterval.

Let then \mu \in \bigcap_n I_n. Since the sequences of \{b_i\}_i and \{a_i\}_i are strictly monotonic, it follows that \mu is none of the extrema. Let \varepsilon>0 be given, and choose n\in \mathbb N such that \frac 1n<\varepsilon. Keep in mind that \mu \in I_n = (a_n,b_n) for any n.

Let \delta>0 such that if |\mu-x|<\delta, then x\in I_n. Since \inf f \leqslant f \leqslant \sup f, for each x\in I_n, it follows that if |x-\mu|<\delta then

\displaystyle |f(x)-f(\mu)|<\frac 1n<\varepsilon,

and thus f is continuous at x=\mu. Since f is integrable over [a,b] it also is integrable over [a,\mu] and [\mu,b]. Set \mu=\mu_1.

By the previous reasoning, we obtain two new different points of continuity, \mu_2 and \mu_3. Having repeated this j times we would obtain 2^{j+1}-1 continuity points. Thus f cannot be continuous over a finite number of points M, for it would mean there exists M such that 2^{j+1}-1<M for each j\in \mathbb N. It follows \Gamma is infinite.

Step 4. \Gamma is dense in [a,b].

Let x\in [a,b] and consider the sequence

\displaystyle\sigma_0=[a,x], \quad \sigma_n=\left[x-\frac{x+a}{2^n},x\right].

The previous argument means there exists for each n a \gamma_n \in \sigma_n with \gamma_n \in \Gamma. But then given \varepsilon>0 we can take obtain an N large enough such that \frac{|x+a|}{2^n}<\varepsilon whenever n\geqslant N and thus

\displaystyle |x-\gamma_n| \leqslant \frac{|x+a|}{2^n}<\varepsilon

for n\geqslant N, that is, \gamma_n \to x. Since x was arbitrarily chosen, \Gamma is dense over [a,b]. This completes the proof.

Source: math.stackexchange


  1. Reblogged this on daodat22121996 and commented:
    Something …. really “dense” 🙂

    Comment by narutomath96 — April 5, 2015 @ 22:54

  2. \displaystyle \sigma_n = \left[x-\frac{x-a}{2^n},x\right]? 🙂

    Comment by Van Hoang Nguyen — December 13, 2016 @ 16:49

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