# Exception to monotone and dominated convergence theorem

If we consider the space $L^\infty ([0,1],\,dx)$, why is it that both monotone and dominated convergence fail?

My first take on the problem was to consider the characteristic function $f_n$ of $[0,1-\frac1n]$, but I wasn't able to proceed from there.

Thank you.

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$f_n$ is increasing to the constant $1$ (hence dominated by it), and converges almost everywhere to this function. Since $|1-f_n|=\mathbf 1_{\left[-\frac 1n,1\right]}$, we have $\lVert 1-f_n\rVert_{\infty}=1$ and we can't have the convergence on $L^{\infty}$-norm. –  Davide Giraudo Nov 21 '11 at 20:49
Is it homework? Did you try what happens to $f_n(x)=x^n$? –  abatkai Nov 21 '11 at 20:49

My suspicion is that you've been assigned the exercise of finding sequences of functions for which the conclusions of those theorem are false, so that $$\int_{[0,1]} \lim_{n\to\infty} f_n \ne \lim_{n\to\infty} \int_{[0,1]} f_n.$$ That does not make these sequences "exceptions" to the theorems, nor does it mean those theorems "fail". The conclusions of those theorems might be said to "fail", but not because the theorems themselves fail. The conclusions of the theorems in fact do not fail in cases where the hypotheses of the theorems hold. Therefore, any instance where the conclusions fail must be an instance where the hypotheses fail. In the case of the dominated convergence theorem, that means there is no dominating function whose integral is finite. In other words, you would look for a case where $$\int_{[0,1]} \sup_n |f_n(x)|\;dx = \infty.$$ If you let $f_n$ be $n$ times the indicator function of $(0,1/n]$, i.e. $$f_n(x) = \begin{cases} n & \text{if } 0 < x\le 1/n, \\ \\ 0 & \text{if }x=0\text{ or }1/n < x \le 1, \end{cases}$$ then $$\lim_{n\to\infty} \int_{[0,1]} f_n(x)\;dx = 1,\text{ and }\int_{[0,1]} \lim_{n\to\infty} f_n(x)\;dx=0$$ as can easily be checked.
Should $f_n(x)$ be $n$ when $0\lt x\leq 1/n$? –  Arturo Magidin Nov 21 '11 at 21:23