It is well known that a function $f:[a,b]\to\mathbb R$ which has the intermediate value property (i.e. if $a',b'\in[a,b]$, then for each $c$ between $f(a')$ and $f(b')$ there is some $x$ between $a'$ and $b'$ such that $f(x)=c$) cannot have jump discontinuities, and that a function $g:[a,b]\to\mathbb R$ of bounded variation can only have jump discontinuities. Therefore, a function having the intermediate value property and being of bounded variation must be continuous. I wonder if the BV-condition can be relaxed a little, so my question is:

Is there a discontinuous function $f:[a,b]\to\mathbb R$ and a function $g:[a,b]\to\mathbb R$ of bounded variation, such that $f$ has the intermediate value property and agrees with $g$ at every point of $[a,b]$ except at countably many?

Any hints/help is highly appreciated. Thanks in advance!

  • $\begingroup$ No, I meant that the function $f$ agrees with a function $g$ nearly everywhere, and $g$ is of bounded variation. I'll reformulate my question, sry for all that confusion. $\endgroup$ – sranthrop Apr 5 '17 at 21:27

Let $x\in[a,b]$ be a discontinuity of $f$. We assume without loss of generality that $x\in(a,b)$, the case of the boundary can be dealt with similarly.

Since $g$ is of bounded variation, the limits $l=\lim_{y\to x^-}g(y)$ and $L=\lim_{y\to x^+}g(y)$ exist and are finite. On the other hand, since $f$ is Darboux and discontinuous at $x$, at least one of $\lim_{y\to x^-}f(y)$ and $\lim_{y\to x^+}f(y)$ must not exist, say the latter.

We hence have that

\begin{align}\forall \epsilon>0,\,&\exists\delta>0,\,\forall y \in(x,x+\delta),\,\,|g(y)-L|<\epsilon\tag{1}\\ \exists \epsilon_0>0,\,&\forall\delta>0,\,\exists y \in(x,x+\delta),\,\,|f(y)-L|\geq\epsilon_0\tag{2}\end{align}

Take $\epsilon=\frac{\epsilon_0}{2}$ in $(1)$, so that there is some $\delta_0>0$ with the property that $|g(y)-L|<\frac{\epsilon_0}{2}$ whenever $y \in (x,x+\delta_0)$. By $(2)$, there is some $y_1\in (x,x+\delta_0)$ with $|f(y_1)-L|\geq \epsilon_0$.

If $g=f$ except on a countable subset of $(x,x+\epsilon)$, there must be some $y_2\in(x,y_1)$ with $g(y_2)=f(y_2)$. Hence, $|g(y_2)-L|=|f(y_2)-L|<\frac{\epsilon_0}{2}$.

Now, since $f$ is Darboux, it must assume all values between $f(y_1)$ and $f(y_2)$ on $(y_1,y_2)$. In particular, it must assume an uncountable range of values $v$ with $\frac{\epsilon_0}{2}\leq |v-L|\leq \epsilon_0$. This means there is some $y_3\in(y_2,y_1)\subset(x,x+\delta_0)$ with $\frac{\epsilon_0}{2}\leq |g(y_3)-L|\leq \epsilon_0$, which contradicts the defining property of $\epsilon_0$. $\square$

  • $\begingroup$ You're welcome! Glad to help. $\endgroup$ – Fimpellizieri Apr 6 '17 at 3:18

A function with IVP agreeing with a BV function except on a countable set must be continuous.

Suppose that $f$ has IVP and agrees with a function $g$ except on a countable set, and that $f$ is discontinuous at some point $c\in[a,b]$. It will eventually be shown that $g$ does not have BV.

It can't be the case that both one-sided limits of $f$ exist at $c$ and equal $f(c)$; WLOG let's say $c>a$ and it's the left-hand side. Then the negation of $\lim\limits_{x\to c-}f(x)=f(c)$ holds. That is, there exists $\varepsilon>0$ such that for all $\delta>0$ there exists $x\in(c-\delta,c)$ with $|f(x)-f(c)|\geq \varepsilon$. For such $x$, either $f(x)\geq f(c)+\varepsilon$, in which case $f((c-\delta,c))$ must contain $(f(c),f(c)+\varepsilon]$, or $f(x)\leq f(c)-\varepsilon$ and $f((c-\delta,c))$ contains $[f(c)-\varepsilon,f(c))$, by the IVP. In either case, after removing any countably infinite set from $(c-\delta,c)$, this still leaves infinitely many $t$ in that interval with $|f(t)-f(c)|>\frac23\varepsilon$, and infinitely many $t$ such that $|f(t)-f(c)|<\frac13\varepsilon$. (The image of a countable set is countable, and intervals in $\mathbb R$ are uncountable.)

Hence the same property must hold for $g$, namely that in every interval of the form $(c-\delta,c)$ there are points $s$ and $t$ such that $|g(s)-f(c)|>\frac23\varepsilon$, and $|g(t)-f(c)|<\frac13\varepsilon$. By taking successively smaller $\delta$, we can arrange sequences of such $s$ and $t$, $(s_n)$ and $(t_n)$, with $s_1<t_1<s_2<t_2<s_3<\cdots$. By taking partitions of $[a,b]$ including arbitrarily many of the $s_n$s and $t_ns$, it can be shown that the total variation of $g$ is larger than any arbitrary multiple of $\frac13\varepsilon$, hence must be infinite.

  • $\begingroup$ Thank you very much. I accepted Fimpellizieri's answer, because I found it a little easier. But this is just my very subjective opinion. $\endgroup$ – sranthrop Apr 6 '17 at 3:01

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.