Sequence of Borel measurable functions where the limit of the integral of the sequence is not equal to the integral of the limit of the sequence

Hello I am trying to think of two different examples of sequences of Borel Measurable functions $$f_n(\omega)$$ where

$$\lim_{n\to\infty} \int\limits_{\Omega} f_n du > \int\limits_{\Omega} (\lim_{n\to\infty} f_n ) du$$

and also

$$\lim_{n\to\infty} \int\limits_{\Omega} f_n du < \int\limits_{\Omega} (\lim_{n\to\infty} f_n ) du$$

I am studying Measure Theory and the book I am using is the Second Edition of Probability and Measure Theory by Robert B. Ash. One of the integration theorems in the book that is related to my question is the Monotone Convergence Theorem which is stated as

"Let $$h_1, h_2, ...$$ form an increasing sequence of Borel measurable functions, and let $$h(\omega) = \lim_{n\to\infty} h_n(\omega), \omega \in \Omega$$ Then $$\int\limits_{\Omega} h_n du \rightarrow \int\limits_{\Omega} h du$$."

This was the direct text stated but I believe what the theorem is trying to conclude is that if $$h_1, h_2 , ...$$ is increasing to some function $$h$$ then $$\int\limits_{\Omega} h_n du \rightarrow \int\limits_{\Omega} h du$$ is the same as $$\lim_{n\to\infty} \int\limits_{\Omega} h_n du = \int\limits_{\Omega} (\lim_{n\to\infty}h_n) du = \int\limits_{\Omega} h du$$

Returning to my question the reason why I want to think of examples of sequences where the integral of the limit the sequence is not equal to the limit of the integral of the sequence (the sequence does not have to satisfy the Monotone Convergence Theorem) is so that I can better understand how the limit of the integral of a sequence of functions is different from the integral of the limit. Any help would be appreciated, thanks.

• $\Omega=(0,1)$ endowed with Lebesgue measure and 1) $f_n := n 1_{(0,1/n)}$ 2) $f_n := - n 1_{(0,1/n)}$. – saz Oct 28 '18 at 20:11

Take a function like $$f(x)=1/(1+x^2)$$. Push that $$n$$ units to the right: $$f_n(x)=f(x-n)$$. Then $$\int f_n\,d\lambda=\pi$$ (Lebesgue measure) but $$\lim_{n\to\infty}f_n(x)=0$$ (pointwise).