Proving that the sum of 2 measurable functions is measurable

[oblixps]

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?

 

[Amer]

To make sure we have a non decreasing sequence
we will abandon some terms to make sure
$$s_n - t_n$$ is a non decreasing

$$f(x) - g(x) > 0$$

take $$\epsilon < \frac{f(x) - g(x) }{2}$$

from the convergence we will have $$n_o \in \mathbb{N}$$
and $$m_o \in \mathbb{N}$$

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

let $$r = max( n_o , m_o )$$
$$s_n - t_n$$ is non decreasing for all $$n > r$$

 

[Opalg]

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]

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]

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)
$$f(x) - g(x) > 2\epsilon$$

since $$t_n$$ is non decreasing, so $$t_n \leq f(x)$$
and $$s_n$$ is non decreasing so $$s_n \leq g(x)$$

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

but $$g(x) > s_n$$
$$0 < t_n - g(x) < t_n - s_n$$ , for n>r

$$t_n \leq t_{n+1} \Rightarrow 0 \leq t_{n+1} -t_n + s_{n+1} - s_{n+1} \leq t_{n+1} - s_{n+1} - t_n + s_n$$ since $$s_n \leq s_{n+1}$$
This
$$0 \leq t_{n+1} - s_{n+1} - ( t_n - s_n )$$ so $$t_n - s_n$$ is non decreasing
Here is a picture for what I did and the epsilon I took
View attachment 1472

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