# A question on limit of integrals of “truncated” functions

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

• What is "$\uparrow$"? – Yanior Weg Oct 19 at 10:40
• It is limit of integrate fn – Narakorn Satwong Oct 19 at 10:50
• Thank you for your reply. I edited four question to avoid further confusion. – Yanior Weg Oct 19 at 11:37

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 $$lim_{n \to \infty} \int_X f_n(x) \mu(dx) = sup\{\int_X f_n(x) \mu(dx)|n \in \mathbb{N}\} = sup\{sup\{\int_X g(x) \mu(dx)|g \text{ is a simple function and } \forall x \in X g(x) \leq f_n(x)\}|n \in \mathbb{N}\} = sup\{\int_X g(x) \mu(dx)|g \text{ is a simple function and } \forall x \in X g(x) \leq f(x)\} = \int_X f_{n}(x) \mu(dx)$$

• thank you so much. – Narakorn Satwong Oct 19 at 14:56

$$0 \leq f_n \leq f$$ 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.

• thank you so much – Narakorn Satwong Oct 19 at 15:01

The sequence $$f_n$$ is increasing.

Let $$x \in X$$.

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

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

By monotone convergence theorem,the statement is true.