Measure theory basic question

[viscid citrus]

This is the second time I've posted this but I still haven't got a satisfactory answer. I have to prove the fact that:
if f:X->[0,infinity] measurable function. Then there exists a sequence of simple functions (function that have a finite elements in the range) (s_n) n>=1 , where s_n:X->[0,infinity) such that:
I) s_n(x) <= s_(n+1)(x), for any x in X (that is to say, that the sequence (s_n) is an increasing sequence of functions)
II) limit of (s_n) is f, for any x in X (pointwise convergence)

Another question I have is if the converse is also true? If a function can be approximated pointwise with simple function, implies that it is a measurable function?

[mortal lion]

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[fossil vault]

What is a measurable function?

[viscid citrus]

A function that returns measurable sets into measurable sets

That is

In this case

f has the propriety that any measurable set in [0, infinity], its preimage through f is also a measurable set in X

[fossil vault]

Yeah

Although that was not entirely what I wanted to know

Let me rephrase that: what are the subsets of R+ that you can measure?

[viscid citrus]

The Lebesgue measurable sets

[fossil vault]

I assume that X is given a suitable sigma-algebra

[fossil vault]

Which is?

Does that correspond to the sigma algebra generated by intervals?

[viscid citrus]

No

The sigma algebra generated by intervals is a subset of the Lebesgue measurable sets

The Lebegue measurable sets are the sets A with the propriety that m(I) = m(A intersect I) + m(A \ I) for any I subset of R ( i think? This is called the Caratheodory criterion)

[fossil vault]

Well, that's good enough then

if you can measure intervals then I think you can already work with it

[viscid citrus]

Ok

Let's work on the borel sigma algebra then

[fossil vault]

You can think of the thing as follows

you take the graph of a measurable function

like a signal

but due to limited precision (not in time, but in values), you are forced to quantize the signal

so for example, instead of outputting pi, you will output 3

so everywhere your signal is between, say, [3, 3.25]

you will underestimate it by using the simple function 3

So basically, if you have a resolution $r(n)$

[wraith fable]

Rion

[fossil vault]

Then in $X_i = f^{-1}([i r(n), (i+1)r(n)])$

[wraith fable]

Rion

[fossil vault]

First of all, $X_i$ is measurable in $X$ by assumption

[wraith fable]

Rion

[fossil vault]

And second, you can use the indicator $i r(n) \chi_{X_i}(x)$

[wraith fable]

Rion

[fossil vault]

Ok the font absolutely sucks

but you get the idea? @viscid citrus

here, I color coded the indicator functions

each color represents a $\chi_{X_i}$

[wraith fable]

Rion

[fossil vault]

and the gap between two horizontal bars is the resolution

[viscid citrus]

What do you mean by resolution function?

r(n)

and why are you multiplying them with i and i+1?

[fossil vault]

for example, if you have a regular subdivision

[viscid citrus]

So you mean to subdivide the range of f?

[fossil vault]

yeah

exactly

[viscid citrus]

How do you know that the range of f is an interval?

[fossil vault]

because you told me so

ah wait

[viscid citrus]

where

[fossil vault]

infinity is included

my bad

[viscid citrus]

Nah

That case can be worked similary

but what if the range of f isn't so nice to work with?

[fossil vault]

well even if it is not finite

it doesn't mean that your subdivision needs to be finite

just that for convenience we want it to be regular

it will help us a lot

[fossil vault]

Can you give me some pathological case you have in mind?

I thought that since $f[X] \subset \mathbb{R}+$ it would suffice to work on $\mathbb{R}+$ like any other interval

[wraith fable]

Rion

[fossil vault]

for infinity, I guess you can also take it separately

Since I assume that ${\infty}$ is also measurable

[wraith fable]

Rion

[viscid citrus]

Yes it is measurable so its fine

[fossil vault]

So without loss of generality we can assume that f(x) is always finite

(f doesn't even have to be bounded)

[fossil vault]

here, to verify the first condition of nondecreasing wrt n

an easy method is to recycle the subdivision from before

so we can divide by 2 the resolution

i.e. take $r(n) = \frac{1}{2^{n}}$

[wraith fable]

Rion

[fossil vault]

it is possible to show that by dividing by 2 the resolution, the simple function approximation only gets better

and cannot decrease at a fixed x

In particular, for a fixed $x$, if $f(x) \in \left[\frac{i}{2^n}, \frac{i+1}{2^n}\right)$

Then $f(x) \in \left[\frac{2i}{2^{n+1}}, \frac{2(i+1)}{2^{n+1}}\right)$

So $f(x)$ is either in $\left[\frac{2i}{2^{n+1}}, \frac{2i+1}{2^{n+1}}\right)$ or $\left[\frac{2i+1}{2^{n+1}}, \frac{2i+2}{2^{n+1}}\right)$

[wraith fable]

Rion

Rion

Rion

[fossil vault]

Case 1: then $s_{n+1}(x) = s_n(x)$ (unchanged underestimation)

[wraith fable]

Rion

[fossil vault]

Case 2: $s_{n+1}(x) = s_n(x) + \frac{1}{2^{n+1}}$ (slightly better underestimation)

[wraith fable]

Rion

[fossil vault]

@viscid citrus

I think this should also allow you to conclude as for why s_n converges pointwise to f

But do tell me if it is not clear

I will do my best to explain again if some parts are not clear

[viscid citrus]

Let me read it

[fossil vault]

Ah, I just noticed that my "simple function" doesn't have a finite range

but we can adjust that part

Denote:
$I_k^{(n)} = \left[ \frac{k}{2^{n}}, \frac{k+1}{2^{n}} \right)$.
Let $X_k^{n} = f^{-1}[I_k^{(n)}]$.
Denote $s_n = \sum_{k=0}^{2^{2n}} \frac{k}{2^{n}} \chi_{X_k^{(n)}}$

[wraith fable]

Rion

[viscid citrus]

Btw, aren't we using the density of the ratioanls in your proof?

[fossil vault]

Of rationals in reals?

[viscid citrus]

Ye

[fossil vault]

Implicitly yes, but we don't need to know that

You can prove that $s_n(x) \leq f(x)$ at any $x$

[wraith fable]

Rion

[fossil vault]

and that the difference is bounded by something that converges to 0

[viscid citrus]

What if we were working with functions that had output in a space without a density propriety, does your proof still work there?

[fossil vault]

That's a good question, I do not think so

since here it is exactly that fragile boundary between countability and continuum

(I'm simplifying measurability to continuum for illustration purposes)

[viscid citrus]

ye

By the way, about the second question I had

If a function has this propriety of being able to be approximated with simple function implies that it is a measurable function?

[fossil vault]

That's a tough one

I do not think that the implication holds

But I am struggling to come up with a counter example

[fossil vault]

I think it is possible to construct such a function on the trivial algebra {empty set, R+}

[viscid citrus]

How so?

[fossil vault]

For instance, if you take a measurable function for the lebesgue measurable sets

you can make a simple function approximation of it

but it may not be measurable in another algebra

I'm not too sure if we still need to ensure that the simple functions are also measurable in the new algebra

No, sorry, I acknowledge the mistake in my ways

https://math.stackexchange.com/questions/1327081/how-to-prove-limit-of-measurable-functions-is-measurable

This should clear it

[viscid citrus]

@fossil vault

Sorry for taking your time again

[fossil vault]

Yo

[viscid citrus]

I think I understand your proof

So we partition the range into intervals and pick the infimum of each partition and with this we create a simple function?

or was this someone else's idea

[viscid citrus]

like in this image I mean

with intervals of of the form [i/2^n, (i+1)/2^n)

[fossil vault]

It was my idea but I assume most textbooks do that

I only picked said resolution because it recycles the previous

[viscid citrus]

Yeah I think everyone had the same idea when they are presented this question

but I am confused what do we do in the case of unbounded functions

Because the range of an unbounded function is ... well unbounded

so when we partition that range again with the idea

[fossil vault]

You use this

[viscid citrus]

we have an infinite sized partition of that range

[fossil vault]

For a given n, you only partition a finite range, which scales with n

[viscid citrus]

Well that is the problem

Wait

[fossil vault]

So for instance

[viscid citrus]

aaaaaaaaaaaaaaaaaa

that's smart

I get it now

[fossil vault]

Cool

[viscid citrus]

So you scale the amount of paritions too

[fossil vault]

I didn't even have to explain

Yep

[viscid citrus]

that's a smart way to get around it

That's all I wanted to know

Thank you for the third time helping me with this question

👍

[fossil vault]

No worries man

[viscid citrus]

Also

You finally start to sound more human

compared ot the other times when we first talked and you used to speak like an AI

xd

Weird compliment

anyway

+close