2
$\begingroup$

Let $f \colon \Omega \to \mathbb{R}$ be measurable and positive almost everywhere.

We set $g(t) := \mu(\{f \geq t\})$.

Let $\varepsilon > 0$ and $k \in \mathbb{N}$ and let's suppose $g(t) < \infty$ for all $t > 0$. We set $$g^\varepsilon := \min(g, g(\varepsilon)), $$

$$f^\varepsilon := f \chi_{\{f \geq \varepsilon\}}$$ and $$f_k^\varepsilon := 2^kf^\varepsilon.$$

Consider the sequence $$ a_k^\varepsilon := 2^{-k} \sum_{n=1}^\infty \mu(\{f^\varepsilon \geq n2^{-k}\}).$$

The author of my book states without proof that $$a_k^\varepsilon \overset{k\to \infty}{\to} \int_0^\infty g^\varepsilon(t) dt.$$

I don't see why this should be trivial. Do you know a proof and can share a hint?

$\endgroup$

1 Answer 1

Reset to default
1
$\begingroup$

Note that $$a_k^\varepsilon = \lim_{i\to \infty}\sum_{n=1}^{2^k i} 2^{-k}g^\varepsilon(n2^{-k})$$ and that these finite sums are right Riemann sums of $g^{\varepsilon}(t)$ corresponding to integral over $[0,i]$ and base length $2^{-k}$. If one is allowed to interchange the limits, then $$\lim_{k\to \infty} a_k^{\varepsilon} = \lim_{i\to \infty} \lim_{k\to \infty}\sum_{n=1}^{2^ki} 2^{-k}g^\varepsilon(n2^{-k}).$$ If the right Riemann sums converge to the Riemann integrals when letting the base length go to zero, we have that $$\lim_{k\to \infty} a_k^{\varepsilon} =\lim_{i\to\infty} \int_0^i g^{\varepsilon}(t) dt = \int_0^\infty g^{\varepsilon}(t) dt.$$

$\endgroup$
10
  • $\begingroup$ Thanks. Interchanging the limits won't be a problem since all summands are positive (Tonelli) but I think your approach doesn't work: Your Riemann sums won't converge to the integral you mentioned since you are not approximating on the interval $[0,i]$ but on $[0,i2^{-k}]$. $\endgroup$
    – el_tenedor
    Sep 18, 2016 at 6:52
  • 1
    $\begingroup$ @el_tenedor You can also write $a_k^\varepsilon=\lim\limits_{i\to\infty} \sum\limits_{n=1}^{2^ki}2^{-k} g^{\varepsilon}(n2^{-k})$, and this will make it work $\endgroup$
    – Del
    Sep 18, 2016 at 8:53
  • $\begingroup$ You are indeed correct, i changed the summation bounds to what @Del wrote. $\endgroup$
    – John
    Sep 18, 2016 at 10:59
  • $\begingroup$ However I'm still not sure about changing the order of limits...in fact we just rewrote the series in another way and magically the limit changed $\endgroup$
    – Del
    Sep 18, 2016 at 14:20
  • $\begingroup$ @Del - It is indeed also a mystery to me, but it seemed that el_tenedor had no problem with this, even though i can't see why Tonelli yields that the limit interchange is allowed. Actually positivity of the summands are not enough, c.f. $anm=(1/n)^{1/m}$ has two different limits. $\endgroup$
    – John
    Sep 19, 2016 at 11:57

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .