A question on limit of integrals of "truncated" functions

Question

Suppose $(X, \Omega, \mu)$ is a measurable space. Suppose $f: X \to \mathbb{R}_+$ is a Lebesgue integrable measurable function. Suppose $f_{n}(x) = \min(f(x),n)$. Is it always true that $$\lim_{n \to \infty} \int_X f_{n}(x)\, \mu(\mathrm{d}x) = \int_X f(x) \,\mu(\mathrm{d}x)?$$

Answer 1

Because $\forall x \in X$ the sequence $f_n(x)$ is monotonously non-decreasing, we can conclude, that $\int_X f_n(x) \mu(dx)$ is also monotonously non-decreasing. Thus $$lim_{n \to \infty} \int_X f_n(x) \mu(dx) = sup\{\int_X f_n(x) \mu(dx)|n \in \mathbb{N}\} = sup\{sup\{\int_X g(x) \mu(dx)|g \text{ is a simple function and } \forall x \in X g(x) \leq f_n(x)\}|n \in \mathbb{N}\} = sup\{\int_X g(x) \mu(dx)|g \text{ is a simple function and } \forall x \in X g(x) \leq f(x)\} = \int_X f_{n}(x) \mu(dx)$$

Answer 2

The sequence $f_n$ is increasing.

Let $x \in X$.

Exists $m \in \Bbb{N}$ such that $f(x)<n,,\forall n \geq m$ thus $f_n(x)=f(x),\forall m \geq n$

So $f_n(x) \to f(x)$

By monotone convergence theorem,the statement is true.

Answer 3

$0 \leq f_n \leq f$ and $f_n \to f$ almost everywhere. Also, $f_n \leq f_{n+1}$. So the result follows immediately from either D CT or monotone convergence theorem.