Today I read monotone convergence theorem , dominated convergence theorem and fatou's lemma
And I need some help
We know the dominated convergence theorem in Measure theory
In its proof we establish a sequence $h_n=2g - |f_n-f|$ the sequence $(h_n)$ of positive measurable function and it converges to $(2g)$ then by fatou's lemma we complete the proof .
The question that presents itself : the sequence $(h_n)$ of positive measurable function and converges to $(2g)$ and I think the sequence $(h_n)$ is increasing because $|f_n-f|$ converges to $0$ So $ |f_1-f| \ge |f_2-f| \ge |f_3-f| \ge |f_4-f| \ge |f_5-f| \ge ....$ then $h_1 \le h_2 \le h_3 \le ...$ so why I don't use monotone convergence theorem to complete the proof .
The sequence $(h_n)$ increasing or not if increasing I need the proof if not I need an example shows the sequence $(h_n)$ is not necessary to be increasing so I can't use monotone convergence theorem
You don't need the sequence $|f_n-f|$ to be decreasing to use the Dominated Convergence Theorem (that's basically the point of using DCT rather than the Monotone Convergence Theorem). Example I thought up on the spot: take an alternating series, $$ 1-x+x^2-\dotsb $$ which you can write as a convergent sequence of partial sums, $$ S_n(x) := \sum_{k=0}^{n} (-x)^{k} = \frac{1-(-x)^{n+1}}{1+x}. $$ These tend pointwise to $\frac{1}{1+x}$ when $|x|<1$. Note the sequence is not uniform, since it is alternately above and below its limiting value. Then the DCT says that $$ \lim_{n \to \infty} \int_0^1 S_n(x) \, dx = \int_0^1 \left( \lim_{n \to \infty} S_n(x) \right) \, dx \\ = \int_0^1 \frac{dx}{1+x} = \left[ \log{(1+x)} \right]_0^1 = \log{2}. $$
(Oh yes, you'll want a dominating function, of course. It is easy to see that $S_n(x) \leqslant 1$ on this interval.)
(Of course, one equality further back is the sum of integrals $$ \sum_{k=0}^{n} (-1)^k \int_0^1 x^k \, dx = \sum_{k=0}^{n-1} \frac{(-1)^{k-1}}{k}, $$ so in this case, the DCT is a sophisticated alternative to the alternating series test that actually gives the value of the sum we all knew.)