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Consider the example $$f_n(x)=\cases{ 0 & if $| x| \ge {1\over n}$\cr n\Bigl(1+nx\Bigr) & if ${-1\over n}\leq x \leq 0 $\cr n\Bigl(1-nx\Bigr) & if $ 0 \le x \le \frac{1}{n}$}$$ Each $f_n$ is piecewise linear and continuous on the interval $[0,1]$, and $f_n(x)\to 1$ as $n\to\infty$ for each $x\in(0,1]$. However, $\int_0^1 f_n(x){\rm d}x=1$ for each $n$, so the conclusion of the Bounded Convergence Theorem does not hold here. So problem should be this sequence of functions is not uniformly bounded?

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    $\begingroup$ The function is a triangle os basis $2/n$ and height $1$, so $\int_0^1 f_n(x)dx= \frac{1\cdot 2/n}{2}=\frac{1}{n}\to 0$ $\endgroup$ Apr 9, 2020 at 17:13
  • $\begingroup$ i was wrong. i correct the sequence $\endgroup$
    – maths
    Apr 9, 2020 at 17:22
  • $\begingroup$ NOw it is NOT uniformly bounded (near 0) because $f_n(0)=n\to\infty$. So YES, that is the problem. $\endgroup$ Apr 9, 2020 at 17:40
  • $\begingroup$ You mean $f_n(x)\to 0$, not $f_n(x) \to 1$, I guess. $\endgroup$
    – PhoemueX
    Apr 9, 2020 at 20:06

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