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Let a real-valued function $f$ be continuous on $[0,1].$ Then there exists a number $a$ such that the integral $$\int_0^1\frac 1 {|f(x)-a|}\, dx $$ diverges. How to prove that statement?

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: whats source? – Maisam Hedyelloo Jun 1 '13 at 10:13
@Maisam Hedyelloo: The human mind. – user64494 Jun 1 '13 at 10:25
There might be two cases of interest. 1) when $f$ vanishes on $(0,1)$; and 2) when it does not. If it does vanish, pick $a=0$ and the integral should diverge. If it does not, then pick $a=f(x_0)$ for some $x_0$ in $(0,1)$? – Cameron Williams Jun 1 '13 at 21:01
Interesting question! I do not know the answer, but a counterexample must be quite pathological. – André Nicolas Jun 2 '13 at 2:43
f(0) and f(1) seem to work. – dimensio1n0 Jun 7 '13 at 2:21
up vote 14 down vote accepted

Take a nondecreasing rearrangement $r(x)$ of the function $f(x)$ (some discussion of this may be found at This involves finding a measure-preserving transformation of the interval $[0,1]$ that transforms $f$ into $r$. In particular, your integrals $\int_0^1\frac 1 {|f(x)-a|}\, dx$ are all preserved (for every $a$). Now apply the result that every monotone function is a.e. differentiable (see Take a point $p$ where the function $r$ is differentiable. Then $a=r(p)$ does the trick, because $r(x)-a$ can be bounded in terms of a linear expression.

Note that the existence of a nondecreasing rearrangement of a function $f$ admits an elegant proof in the context of its hyperreal extension $^\star f$, which we will continue to denote by $f$. Namely, take an infinite hypernatural $H$ and consider a partition of the hyperreal interval $[0,1]$ into $H$ segments, by means of partition points $0, \frac{1}{H}, \frac{2}{H}, \frac{3}{H}, \ldots, \frac{H-1}{H}, 1$. Now rearrange the values $f(\frac{i}{H})$ of the function at partition points in increasing order, and permute the $H$ segments accordingly. The standard part of the resulting function is the desired monotone function $r$.

Note 1. I should point out that one does not really need to use the result that monotone functions are a.e. differentiable. Consider the convex hull of the graph of $r(x)$, and take a point where the graph touches the boundary of the convex hull (other than the endpoints 0 and 1). Setting $a$ equal to the $x$-coordinate of the point does the job.

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What if the derivative is 0 at every differentiator point, as in the Lebesgue function of a cantor set? – Brian Rushton Jun 6 '13 at 14:14
The constant functions are obvious counterexamples, so to make the problem meaningful in this case one could declare that the integral $\int_0^1 \frac{1}{0}dx$ "diverges". – Mikhail Katz Jun 6 '13 at 14:32
There are strictly monotonic functions whose derivative is 0 at every point where it is defined. – Brian Rushton Jun 6 '13 at 14:33
If one lets $a$ be the value of the function at that point, the integral $\int \frac{1}{f(x)-a}dx$ will indeed diverge. – Mikhail Katz Jun 6 '13 at 14:38
True! That's a good point... – Brian Rushton Jun 6 '13 at 14:45

Here's a proof that $\int_0^1\frac1{\lvert f(x)-a\rvert}dx = \infty$ for every $a$ in the image of $f$ and outside of a meagre set. In particular, if $f$ is not constant, then there are uncountably many such $a$ in every neighborhood of the image of $f$. [Note: I also agree that user72694's proof works fine, and is completely independent of the proof I'll give here.]

First, define $g(a)$ to be the given integral, and let $[a_0,a_1]$ be the image of $f$. Assuming that $f$ is non-constant, we have $a_0 < a_1$. Letting $b_0 < b_1$ be in $[a_0,a_1]$ then Fubini's theorem gives, $$ \int_{b_0}^{b_1}g(a)\,da= \int_{b_0}^{b_1}\!\!\!\int_0^1\frac1{\lvert f(x)-a\rvert}\,dxda =\int_0^1\!\!\!\int_{b_0}^{b_1}\frac1{\lvert f(x)-a\rvert}\,dadx=\infty. $$ Here, the integral of $1/\lvert f(x)-a\rvert$ wrt $a$ is infinite whenever $f(x)$ is in $[b_0,b_1]$ (because $1/x$ is not integrable at the origin), which happens for $x$ in a nontrivial interval, so the double integral is infinite. This means that $g$ is not integrable (and, hence, is unbounded) in any nontrivial interval $[b_0,b_1]$ in the image of $f$.

Next, for each $K > 0$, set $S_K=\lbrace a\in[a_0,a_1]\colon g(a) > K\rbrace$. As $g$ is unbounded in the neighborhood of any point, the set $S_K$ is dense in $[a_0,a_1]$. Furthermore, $S_K$ is open for each positive $K$. To see this, note that $g_n(a)\equiv\int_0^11_{\lbrace\lvert f(x)-a\rvert > 1/n\rbrace}/\lvert f(x)-a\rvert dx$ is a sequence of continuous functions increasing to $g$, so $g$ is lower semicontinuous.

Hence, we have $$ \left\lbrace a\in[a_0,a_1]\colon g(a)=\infty\right\rbrace=\bigcap_{n=1}^\infty S_n, $$ which is a countable intersection of dense open sets in $[a_0,a_1]$ so, by definition, its complement is meagre.

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Very nice!${}{}$ – Nick Strehlke Jun 6 '13 at 20:35
Both answers to the question under consideration, which belongs to mathematical folklore and is authored by S. Konyagin, are right and nice. Unfortunately, I cannot share bounty between the two answers. – user64494 Jun 7 '13 at 4:55
That's interesting. His wiki page that you linked does not say anything about the question. Can you elaborate? – Mikhail Katz Jun 7 '13 at 7:22

I don't think this statemnt is correct if $a\in[0,1]$. Maybe I'm wrong but you can take the counter example $f(x)=\frac 1 {|sin(x)+cos(x)|}$. So let's make two points: 1.looking at $|f(x)-a|$,notice that $f(x)=sin(x)+cos(x)$ is continious but moreover is nonegative. if $a\in[0,1]$ the integral is well defined. 2.Then the antriderivative of the function is wel defined according to maple and for all a in [0,1] the claim is correct.

EDIT: Let's try reductio ad absurdum and assume $\int_0^1 \frac 1 {|f(x)-a|} dx$ converges to $L\in\mathbb R$. Also let $a\in \mathbb R,f:[0,1]\rightarrow\mathbb R$. since the function is defined on a closed interval $\exists x_0 \in \mathbb R$ s.t $x_0$ is a maximum (concluding from weirestrass first lemma for continious function on closed intervals) . since $|f(x)-a|\leq |f(x_0)-a| \Longrightarrow \frac {1}{|f(x)-a|}\geq \frac {1}{|f(x_0)-a|}$. and taking $a\to f(x_0)$, $|f(x_0)-a|\to 0$ but then $ \frac 1 {|f(x_0)-a|} \to \infty$ and finally $L=\int_0^1 \frac 1 {|f(x)-a|}dx\geq \int_0^1 \frac 1 {|f(x_0)-a|}\to \infty$ which is undifined in contradiction for that we assumed that $L\in\mathbb R$.

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Thank you for the interest to the question. Unfortunately, you did miss the point. The statement does not assert that $a \in [0,1].$ – user64494 Jun 1 '13 at 10:04
I edited the post. – user65985 Jun 1 '13 at 10:32
Mmh i think there are some flaws in the proof (hope,that I'm not wrong) 1) You can't assert that exists $x_0$ s.t. $f'(x_0)=0$ only because f is defined on a close, maxima can be reach on the extremis of the interval and therefore the derivate is always $\neq 0$. Just take as counterexample $f(x)=x-\frac{2}{3}$. 2) the disequality about absolute value is not true in general, consider the previous $f$, if you take the maxima $=\frac{1}{3}$ and $a>0$ and a random point in $[0,\frac{2}{3}]$ you obtain a contradiction. – Riccardo Jun 1 '13 at 21:11
a. the derivative does not matter for the question. The important thing is you have a maximum (extreme value therom which is also explained here b. I don't think you can take the maximum from $[0, \frac 2 3]$ since you choose $x_0$ from $[\frac 2 3,1]$. if f is non-negtive the claim is correct. for the counter-example you wrote since the claim is correct in [\frac 2 3, 1],since the $\int_{\frac 2 3}^{1}$ diverges the whole integral divereges. For negitve function take $x_1$, the minimum of the function in almost a same way. – user65985 Jun 2 '13 at 2:28
If $x_0$ is a maximum for $|f(x)-a|$ then $a$ is (by implicit assumption) fixed, so you can't make $a\to f(x_0)$ and expect the inequality $|f(x) - a|\leq |f(x_0)-a|$ to continue to hold for all $x\in [0,1]$. – Nick Strehlke Jun 2 '13 at 2:42

The Devil's staircase is a counterexample. When the integral is defined, it converges. For instance, at 0, the Devil's staircase is approximately between $(C_1 x)^{\frac{\ln 2}{\ln 3}}$ and $(C_2 x)^{\frac{\ln 2}{\ln 3}}$ (the curves connecting the left endpoints of intervals and right endpoints, respectively), and so your integral converges if $a=0$. Near any other point where the integral is defined, the integral converges since the set is self-similar, so all such points are like 0.

All of this is in Chapter four of Frank Jones' Lebesgue integration book, as well as pages 521.

Devil's staircase from Wikipedia

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Good point! I'll have to think of a way to address that. – Brian Rushton Jun 2 '13 at 2:47
I believe this works now. – Brian Rushton Jun 6 '13 at 14:30
The constant functions are obvious counterexamples; so to make the problem meaningful in this case one could declare that the integral $\int_E \frac 1 0 dx$ over the set $E$ of positive measure diverges. This is a usual convention in real analysis. – user64494 Jun 6 '13 at 15:09

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