About a measurable function in $\mathbb{R}$

Solution 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 $[2^{m-1}, 2^m]$, $|\intop_I h \, dx| \le 2^{m/2} |I| \le |I|^{1/2}$.