Talk:Simple function

Stardard representation

It would be nice to include a section on the standard representation of simple measurable functions. If I have time I'll get to it.

Goldencako 19:15, 13 January 2011 (UTC)

Confusing sentence

We set $I_{n,k}=\left[{\frac {k-1}{2^{n}}},{\frac {k}{2^{n}}}\right)$ for $k=1,2,\ldots ,2^{2n}$ and $I_{n,2^{2n}+1}=[2^{n},\infty ]$.

Someone should explain how it is that $I_{n,2^{2n}+1}=[2^{n},\infty ]$. I don't think it is correct.

Timhoooey 00:12, 8 October 2007 (UTC)

That's the definition of $I_{n,2^{2n}+1}$, as indicated by the 'we set'. Algebraist 11:06, 1 June 2008 (UTC)

An explanation of the confusing sentence:

$I_{n,2^{2n}+1}$ is the $2^{2n}+1th$ segment of the partition we are creating on the range (y) of the function f. Since the $2^{2n}+1th$ segment has to cover the entire remaining range, we extend it from $2^{2n}/2^{n}=2^{n}$ to $\infty$ (viz. ${\frac {k-1}{2^{n}}},k=2^{2n}+1)$. I know approximating $[2^{n},\infty ]$ by $2^{n}$ seems odd, but everything will work out when we take the limit.

So, we let n approach infinity. Then our estimate improves dramatically. In fact, as n approaches $\infty$, the limit of the sequence of functions, $f_{n}$, is equal pointwise to the function f. ($1/2^{n}$ > |f(x₀) – f_{n}(x₀)| ≥ 0, for each $x=x$₀) Notice that our $2^{2n}+1th$ segment now covers just one point: $\infty$.

--129.71.204.146 (talk) 19:17, 4 December 2009 (UTC)

Proof that f is increasing

Let $f(x$₀$)=y$₀. Then y₀ is in $[(k-1)/2^{n},k/2^{n}),$ for some k = 1, ..., $2^{2n},$ or else y₀ is in $[2^{n},\infty ]$. Therefore $f_{n}(x$₀$)=(k-1)/2^{n}$ or $f_{n}(x$₀$)=2^{n}$.

For n+1, y₀ is in $[(k-1)/2^{n},(2k-1)/2^{n+1})\cup [(2k-1)/2^{n+1},k/2^{n})$--since we have divided the original intervals by two and added another ${\frac {1}{2^{n}}}$th--or else y₀ is in $[2^{n+1},\infty ]$. So $f_{n+1}(x$₀) is in the set {$(k-1)/2^{n}$, $(2k-1)/2^{n+1}$, $2^{n+1}$}. Matching the ranges of $f_{n}$ and $f_{n+1}$ we see that in all cases, $f_{n+1}(x$₀$)$≥$f_{n}(x$₀$)$. Therefore, $f_{n}(x)$ is an increasing function in n.

--129.71.204.146 (talk) 19:17, 4 December 2009 (UTC)

all step functions are simple

Isn't this at odds with the requirement of finite codomain? I would think of the floor function as being a step function but it takes on infinity values. Crasshopper (talk) 21:48, 19 October 2014 (UTC)

Great writing

The opening paragraph of this article is what the rest of math wikipedia should aspire to. Crasshopper (talk) 21:49, 19 October 2014 (UTC)

Reconcile finite with countable number of values?

Can we reconcile or refine the definition of simple function in this article with the defintion given by the Springer online encyclopedia of mathematics, which says a simple function is a measureable function that takes on at most a countable number of values? http://www.encyclopediaofmath.org/index.php/Lebesgue_integral I think the analysis book by Kolmogorov and Fomin uses the same definition as Springer

Tashiro (talk) — Preceding undated comment added 17:50, 10 January 2015 (UTC)

Countably valued functions - why?

Based on WP:ONUS, this generalization seems obvious and therefore does not improve the article: of course simple functions may be generalized to have countable ranges. The article on Bochner measurable functions does a good-enough job briefly introducing countably valued functions in the intro section, so linking to this article is unhelpful. Linking to the BMF in "See also" section would be more appropriate. StrokeOfMidnight (talk) 01:53, 30 January 2023 (UTC)

@StrokeOfMidnight: the point is, it's not about being an obvious generalization but rather about what the term countably valued function means. For the reader it's certainly not obvious that this term is explained in the article Bochner measurable. And I don't think the article Bochner measurable does a good job because it uses the term before it explains what it is. This is bad style.--Tensorproduct (talk) 08:20, 2 February 2023 (UTC)

This term is not essential even for the definition of BMF, for which it's intended. Mentioning "countably-valued" just before the formula $f(t)=\ldots$ can be skipped without affecting clarity. Therefore the term is not important enough to be discussed in the current article. StrokeOfMidnight (talk) 02:11, 3 February 2023 (UTC)

It's not important whether the term is essential or not. I mean this is also not for us to decide and one can use the term also for other things than the definition of Bochner measurability. How about a compromis, that in the article simple function it will be mentioned shortly in one sentence, that the function extended to countably values is called countably valued function? I am not sure I understand, why you are so much against it.--Tensorproduct (talk) 08:31, 3 February 2023 (UTC)

@StrokeOfMidnight: I didn't get any answer.--Tensorproduct (talk) 18:26, 16 February 2023 (UTC)

Gyus? Have you heard about the policies WP:RS/WP:CITE? The article and the disputed issue are unverifiable by an independent wikipedian. Lokys dar Vienas (talk) 02:23, 23 February 2023 (UTC)

If the issue is only a source, I can provide one. But for StrokeOfMidnight that is not the issue, the user just doesn't want the concept to be on Wikipedia because "it is obvious". I tried to discuss, but I waited 3 weeks and didn't get an answer.--Tensorproduct (talk) 08:51, 23 February 2023 (UTC)

It does not matter whether it is obvious or not. It is not supplied by reference which relates the two concepts. We can base and judge article text only by the published reliable sources. I strongly recommend you to carefully read and understand the policies I mentioned above. Without references you are wasting other wikipedians' time. Lokys dar Vienas (talk) 18:13, 23 February 2023 (UTC)

Go on and delete it, I don't care anymore and have no interest to look for a source anymore, if you talk to me like that. ("..waste other Wikipedians time.."). I also waste a lot of time.--Tensorproduct (talk) 19:14, 23 February 2023 (UTC)

I still want the term too be displayed. It's an important term--Tensorproduct (talk) 01:27, 4 March 2023 (UTC)