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Proving that the sum of 2 measurable functions is measurable

oblixps

Member
May 20, 2012
38
I know there are many proofs for this but I am having trouble proving this fact using my book's definition.

My book defines first a non negative measurable function f as a function that can be written as the limit of a non decreasing sequence of non-negative simple functions.

Then my book defines that a function (taking on both positive and negative values) is measurable if both its positive part and negative part are measurable.

Let f and g be 2 measurable functions. Show that f + g is meaasurable as well.
If f and g are both non-negative, then it is clear that if $s_n$ is a nondecreasing sequence converging to f and $t_n$ is a nondecreasing sequence converging to g, then $s_n + t_n$ is a nondecreasing sequence converging to f + g.

However, I am having trouble with the more general case where f and g can take both positive and negative values. I am trying to show that the positive part of (f + g) is measurable, which means there exists a nondecreasing sequence of non-negative simple functions converging to $(f + g)^+$.

If I choose an x where f(x) and g(x) are non-negative, then it is clear how to construct such a sequence.

However, if I choose an x where f(x) > 0 and g(x) < 0 and f(x) + g(x) > 0, then I can represent this as $f^+ - g^-$. but i can't seem to come up with a sequence of functions converging to $f^+ - g^-$ AND is non decreasing. If $s_n$ is a nondecreasing sequence converging to $f^+$ and $t_n$ is a nondecreasing sequence converging to $g^-$, $s_n - t_n$ converges to $f^+ - g^-$ but it may not be a non decreasing sequence.

Can someone help me with this problem?
 
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Amer

Active member
Mar 1, 2012
275
To make sure we have a non decreasing sequence
we will abandon some terms to make sure
[tex] s_n - t_n [/tex] is a non decreasing

[tex] f(x) - g(x) > 0 [/tex]

take [tex] \epsilon < \frac{f(x) - g(x) }{2} [/tex]

from the convergence we will have [tex] n_o \in \mathbb{N} [/tex]
and [tex] m_o \in \mathbb{N}[/tex]

such that [tex] \mid f_n - f(x) \mid < \epsilon [/tex] for all [tex] n > n_o [/tex]
and [tex] \mid g_n - g(x) \mid < \epsilon [/tex] for all [tex] n >m_o [/tex]

let [tex] r = max( n_o , m_o ) [/tex]
[tex] s_n - t_n [/tex] is non decreasing for all [tex] n > r [/tex]
 

Opalg

MHB Oldtimer
Staff member
Feb 7, 2012
2,725
I know there are many proofs for this but I am having trouble proving this fact using my book's definition.

My book defines first a non negative measurable function f as a function that can be written as the limit of a non decreasing sequence of non-negative simple functions.

Then my book defines that a function (taking on both positive and negative values) is measurable if both its positive part and negative part are measurable.

Let f and g be 2 measurable functions. Show that f + g is measurable as well.
That is a terrible way to define a measurable function. The usual approach is to define measurable functions in terms of measurable sets. Then a function $f$ is said to be measurable if for every real number $\alpha$ the set $\{x:f(x)>\alpha\}$ is measurable. In that case, if $f$ and $g$ are measurable functions then the set $\{x:f(x) + g(x)>\alpha\}$ is the union of the sets $\{x:f(x)>r\}\cap \{x:g(x)>\alpha - r\}$ as $r$ runs through the rational numbers. That is a countable union of measurable sets and is therefore measurable.

Starting from your book's definition of measurability, it does not even seem easy to show that the sum of a measurable function and a constant function is measurable.
 

oblixps

Member
May 20, 2012
38
Amer, could you explain a little more why $s_n - t_n$ must be non decreasing for n > r?
I don't really see how that follows from knowing $|s_n(x) - f(x)| < \epsilon$ and $|t_n(x) - f(x)| < \epsilon$. These expressions tell us how to bound $s_n - t_n$ for each n, but doesn't seem to give us a relationship between the nth and (n+1)th term.

Opalg, I agree that the more standard definition of a measurable function makes this problem much easier to prove.

thank you for both of your answers!
 

Amer

Active member
Mar 1, 2012
275
Amer, could you explain a little more why $s_n - t_n$ must be non decreasing for n > r?
I don't really see how that follows from knowing $|s_n(x) - f(x)| < \epsilon$ and $|t_n(x) - f(x)| < \epsilon$. These expressions tell us how to bound $s_n - t_n$ for each n, but doesn't seem to give us a relationship between the nth and (n+1)th term.

Opalg, I agree that the more standard definition of a measurable function makes this problem much easier to prove.

thank you for both of your answers!
f(x) > g(x)
[tex] f(x) - g(x) > 2\epsilon [/tex]

since [tex] t_n [/tex] is non decreasing, so [tex] t_n \leq f(x) [/tex]
and [tex] s_n [/tex] is non decreasing so [tex] s_n \leq g(x)[/tex]

for n>r
[tex] \mid f(x) - t_n \mid < \frac{f(x) - g(x) }{2} [/tex]
we can take it without the absolute value since f(x) > tn
[tex] f(x) - t_n < \frac{f(x) - g(x) }{2} < f(x) - g(x) [/tex]
[tex] f(x) - t_n < f(x) - g(x) [/tex]
[tex] 0 < t_n - g(x) [/tex] for n>r

but [tex] g(x) > s_n [/tex]
[tex] 0 < t_n - g(x) < t_n - s_n [/tex] , for n>r

[tex] t_n \leq t_{n+1} \Rightarrow 0 \leq t_{n+1} -t_n + s_{n+1} - s_{n+1} \leq t_{n+1} - s_{n+1} - t_n + s_n [/tex] since [tex]s_n \leq s_{n+1} [/tex]
This
[tex] 0 \leq t_{n+1} - s_{n+1} - ( t_n - s_n ) [/tex] so [tex] t_n - s_n [/tex] is non decreasing
Here is a picture for what I did and the epsilon I took
Untitled.jpg

I hope of solution is clear now
Edited: :(
I made a mistake sorry, if we have [tex] t_{n+1} = t_n [/tex]
and [tex] s_{n+1} > s_n [/tex]
we get [tex] t_{n+1} - s_{n+1} < t_n - s_n [/tex]
Thinking of something else
 
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