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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 $$\lim_{n \to \infty} \int_X f_{n}(x)\, \mu(\mathrm{d}x) = \int_X f(x) \,\mu(\mathrm{d}x)?$$

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  • $\begingroup$ What is "$\uparrow$"? $\endgroup$ – Yanior Weg Oct 19 at 10:40
  • $\begingroup$ It is limit of integrate fn $\endgroup$ – Narakorn Satwong Oct 19 at 10:50
  • $\begingroup$ Thank you for your reply. I edited four question to avoid further confusion. $\endgroup$ – Yanior Weg Oct 19 at 11:37
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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 $$lim_{n \to \infty} \int_X f_n(x) \mu(dx) = sup\{\int_X f_n(x) \mu(dx)|n \in \mathbb{N}\} = sup\{sup\{\int_X g(x) \mu(dx)|g \text{ is a simple function and } \forall x \in X g(x) \leq f_n(x)\}|n \in \mathbb{N}\} = sup\{\int_X g(x) \mu(dx)|g \text{ is a simple function and } \forall x \in X g(x) \leq f(x)\} = \int_X f_{n}(x) \mu(dx)$$

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  • $\begingroup$ thank you so much. $\endgroup$ – Narakorn Satwong Oct 19 at 14:56
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$0 \leq f_n \leq f$ 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.

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  • $\begingroup$ thank you so much $\endgroup$ – Narakorn Satwong Oct 19 at 15:01
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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.

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