Let $h:\mathbb{R}\rightarrow\mathbb{R}$ be a measurable function such that

$$\left|\int_I h\right|\leq c \sqrt{|I|}$$

for each interval $I$. Then $h_\epsilon(x)=h(x/\epsilon)$ satisfies

$$\int_Ah_\epsilon(x)dx\rightarrow0 $$ as $\epsilon$ goes to zero, for each Borel set $A$ such that $|A|<\infty$.

  • 3
    $\begingroup$ Do a change of variables. $\endgroup$ Commented Mar 2, 2012 at 10:15
  • $\begingroup$ I edited the question based on chessmath's comments on a currently deleted answer. $\endgroup$ Commented Mar 2, 2012 at 11:15
  • $\begingroup$ Could you provide an example of $f$ which satisfy the hypothesis but which is not square integrable? $\endgroup$ Commented Jul 26, 2012 at 19:50
  • $\begingroup$ @DavideGiraudo $x^{-1/2}$ $\endgroup$ Commented Oct 3, 2012 at 0:55
  • $\begingroup$ What are the precise integrability assumptions on $h$? $L^1$, $L^1_\mathrm{loc}$, or what? $\endgroup$ Commented Oct 3, 2012 at 1:04

1 Answer 1


The answer depends on the integrability assumptions on $h$:

  1. If $h \in L^1$, or indeed $L^q,q<\infty$, this is trivial, since $\Vert h_\epsilon \Vert \to 0$.

  2. If $h \in L^\infty$, then this is true, since this holds for "elementary" sets, that is, fininte unions of intervals, and hence for any other sets by means of $L^1$ approximation of their indicators.

  3. If $h \in L^1_\mathrm{loc}$, this is just not true, and I have a counterexample.

So here it is. Let's denote

$s(x) := \begin{cases} 1, & x\in[2k,2k+1)\\ -1, & x\in[2k+1,2k+2) \end{cases}$

$h(x):=\begin{cases} 0, & x\le 1\\ 2^{n/2}s(2^{n}x), & x\in(2^{n},2^{n+1}] \end{cases}$

I will have to show that it satisfies the square-root hypothesis later, but first let's construct a set $A$ and see why it violates the claim.

$A := \bigsqcup_n A_n$, where $A_n := \bigcup_k [2^{-2n^2} \cdot 2k, 2^{-2n^2} \cdot (2k+1)] \cap [1+2^{-n-1}, 1+2^{-n})$.

Then $h(2^{n^2} x) = 2^{n^2/2} s(2^{2n^2} x), x \in [1,2]$, so it will only "resonate" with $A_n$:

$\displaystyle \intop_A h_{2^{-n^2}} dx = \intop_{A_n} h_{2^{-n^2}} dx$.

The latter integral equals $2^{n^2/2-n}$ up to a constant.

Finally, let's estimate $\intop_I h \, dx$. Clearly, we may assume that $I \subset [0,+\infty)$. I claim that if $[2^m,2^n] \subset I \subset [2^{m-1},2^{n+1}]$, then $|\intop_I h \, dx| \le 2^{-m/2}$, up to a constant, since an integral over $[2^m,2^n]$ is zero, and when changing the left or right limit of integration within the abovementioned segments the integral varies within $2^{-m/2} + 2^{-n/2}$. So for sufficiently long segments (that is, longer than $2^{-m}$) everything is fine. For the ones shorter than $2^{-m}$, within $[2^{m-1}, 2^m]$, $|\intop_I h \, dx| \le 2^{m/2} |I| \le |I|^{1/2}$.

  • 1
    $\begingroup$ Nice counter-example! You deserved the bounty. $\endgroup$ Commented Oct 6, 2012 at 10:01

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