0
$\begingroup$

A simple function $\varphi$ is a finite sum of the form

$$\varphi (x) := \sum^{N}_{k = 1} r_k \mathbf{1}_{E_{k}} (x)$$

where $E_k$ is measurable for every $k \in \{1, \dots , N\}$ and $\{r_k\}^{N}_{k=1} \subseteq \mathbb{R}$, with $\mathbf{1}$ being the indicator (or characteristic) function.

Stein & Shakarchi (at page 50 of their book) define the canonical form of $\varphi$ as the unique decomposition as above, where the real numbers $r_k$ are distinct and non-zero and the sets $E_k$ are disjoint (and measurable), for every $k \in \{1, \dots , N\}$. Then, with respect to the relation between the canonical form as they define it and the Lebesgue integral, show in their Proposition 1.1 at page 51 that the limitation of focusing on the canonical form as they define it is apparent.

On the contrary, for example, Royden & Fitzpatrick do not ask the real numbers $r_k$ to be non-zero, for every $k \in \{1, \dots , N\}$

Now, personally I find the definition in Royden & Fitzpatrick more natural. Thus, I was wondering why there is the need to impose the constant real numbers to be different from $0$ in Stein & Shakarchi in the very first place.

Any feedback would be most welcome!

$\endgroup$

2 Answers 2

Reset to default
3
+25
$\begingroup$

It seems to me that the first definition adds the restriction that $r_k$ are non-zero to make the decomposition unique. If, say, $r_{N+1}$ is allowed to be $0$, one could add it to the sum or discard it, without changing the function, thus making the decomposition not unique.

$\endgroup$
2
  • $\begingroup$ Thanks a lot for the answer. Actually, I thought something along these lines: as long as we are explicit (and of course we are) on the domain $D$ of $\varphi$, if there is a measurable set $E_{n+1} \subseteq D$ where $\varphi$ takes value $r_{n+1} = 0$, then we can dismiss it from the representation in the canonical form, because it does not change anything (as a sum). However, I also felt – maybe wrongly – that this was not really hampering the uniqueness of the representation. Hence, my question. $\endgroup$
    – Kolmin
    Sep 18 at 11:36
  • $\begingroup$ Once more, thanks a lot for the answer and let's see if other takes on the issue are going to come up! $\endgroup$
    – Kolmin
    Sep 18 at 11:36
0
$\begingroup$

I think that this restriction it for convenience: if you endow the ambient space with some infinite measure, you just have to say that all the $E_k$ must have finite measure for $\varphi$ to be integrable.

$\endgroup$
5
  • $\begingroup$ Thanks for the answer. However, I do not really see how the idea of infinite measure is related to ruling out the possibility that there exists a measurable set where the function takes value $0$. $\endgroup$
    – Kolmin
    Sep 18 at 7:47
  • $\begingroup$ If one $r_k$ may vanish, the condition for integrability is less nice: $\varphi$ is integrable if and only if the $E_k$ have finite measure for all $k$ such that $r_k \ne 0$. $\endgroup$
    – Christophe Leuridan
    Sep 18 at 12:08
  • $\begingroup$ But why? You don't need the additional condition "$r_K \neq 0$", because even if we have a $k$ such that $r_k = 0$, that still does not play any role in the integral (as it should be – the function takes value $0$ there). $\endgroup$
    – Kolmin
    Sep 18 at 19:33
  • 1
    $\begingroup$ Thus, if $r_k=0$, the finiteness of the measure of $E_k$ is not required for $\varphi$ to be integrable. $\endgroup$
    – Christophe Leuridan
    Sep 19 at 11:34
  • $\begingroup$ Thanks for the clarification (sorry for being late). $\endgroup$
    – Kolmin
    Sep 24 at 9:20

You must log in to answer this question.

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