Tag: Analysis

This problem comes from Bass. I have a problem with my proof that I could not solve it. It may not even be the right way to solve it, but that's what I proposed.

Right here, $ int $ is equivalent to $ int_ mathbb {R} $

Problem:

Let $ (X, mathcal {A}, mu) $ to be a measurement space. Suppose you have the functions $ f_n $ and $ f $ which are integrable and not negative (for all $ n $ in the case of $ f_n $). In addition, suppose that $ f_n to f $ almost everywhere, and $ int f_n to int f $. Prove that for every $ A in mathcal {A} $.

$$ int_Af_n mathrm {d} mu to int_A f mathrm {d} mu $$

My attempt:

We can rewrite the integral of $ f_n $ as

$$ int_A f_n mathrm {d} mu = int f_n chi_A mathrm {d} mu $$
or $ chi_A $ is the indicator function for the whole $ A $.

For everyone $ n $, define the functions

$$ g_n (x): = f_n (x) chi_A (x) =
begin {cases}
f_n (x), & x in A \
0, & x notin A
end {cases}
$$
For each $ n $, $ g_n (x) geq 0 $, so $ | g_n (x) | = g_n (x) leq f_n (x) $

Now we get the limit

$$ lim_ {n to infty} g_n = lim_ {n to infty} (f_n (x) chi_A (x)) = chi_A (x) lim_ {n to infty} f_n ( x) = f (x) chi_A (x) $$ for everyone $ x $ such as $ f_n (x) to f (x) $.

Since $ f_n to f $ almost everywhere, it follows that $ g_n to g = f chi_A $ almost everywhere.

Moreover, by hypothesis, since each $ f_n $ is integrable, we get

$$ g_n leq f_n text {a.e.} Rightarrow | g_n | leq | f_n | text {a.e.} Rightarrow int | g_n | mathrm {d} mu leq int | f_n | mathrm {d} mu < infty $$

So for all $ n $ we have that $ g_n $ is integrable.

Functions $ g_n $ are measurable since the function of the indicator and $ f_n $ are measurable. Therefore, the multiplication of the two functions is also measurable.

Therefore, we found that $ g_n $ are measurable, and that $ g_n to g $ almost everywhere. It means that I have almost all conditions (without wanting to play pun) of the conditions required to use the dominated convergence theorem. I miss finding an embeddable feature $ h: X to (0, infty) $ such as, $ | g_n (x) | leq h (x) $ A.E.

The closest I got to this was the previous statement in this answer that $ | g_n (x) | = g_n (x) leq f_n (x) $ A.E.

I do not know how to find a function that absolutely links $ g_n $ and does not depend on $ n $and, since we have not assumed $ f_1 leq f_2 leq … $ I can not say that $ f_n leq f $ which would solve this problem. What do I miss?

(Finally, I will finish the proof assuming that I found such a function $ h $ this allows me to use the DCT)

If we invoke the DCT, we have that

$$ lim_ {n to infty} int_A f_n mathrm {d} mu = lim_ {n to infty} int f_n chi_A mathrm {d} mu = lim_ {n to infty} int gnn mathrm {d} mu = int lim_ {n to infty} g_n mathrm {d} mu = int g mathrm {d} mu = int f chi_A mathrm {d} mu = int_A f mathrm {d} mu $$