- $\displaystyle \int_0^1 \lim_{n\to\infty} f_n(x)\,dx = \lim_{n\to\infty}\int_0^1f_n(x)\,dx $
- There is no function $\,g:\left[0,1\right]\to \mathbb R\,$ lebesgue integrable such that $\,\left\lvert f_n (x)\right\rvert\le g(x)\,$ for $\,0\le x\le 1\, $ and $\, n\ge 1 \,$
2026-05-05 07:58:54.1777967934
existence of a sequence of continuous functions with two conditions
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Take the sequence: $$f_n(x) = \begin{cases} n\qquad \mbox{if }\; \frac{1}{n+1}\leq x< \frac{1}{n} \\ 0 \qquad \mbox{otherwise}\end{cases}$$ We have $\int_{0}^1 f_n(x) dx = \frac{1}{n+1}$. So $\int_0^1 f_n(x) dx \to 0$. What happens to $g(x)=\max\{f_n(x): n\in \mathbb{N}\}$?
Now you need to smoothen $f_n$ a little bit, to make them continuous...