From the text Real Analysis by Royden, he gives the following proposition
Proposition 8: Let $\{f_n\}$ be a sequence of bounded measurable functions on a set of finite
measure $E$.
$$\text{If } \{f_n\}\to f \text{ uniformly on $E$, then } \lim_{n\to\infty}\int_{E}f_n = \int_{E}f.$$
The proof is given as:
Proof: Since the convergence is uniform and each $f_n$ is bounded, the limit function $f$ is
bounded. The function $f$ is measurable since it is the pointwise limit of a sequence of
measurable functions. Let $\epsilon>0$. Choose an index $N$ for which
$$\left|f-f_n\right|<\epsilon/\operatorname{m}\left(E\right)\text{ on $E$ for all $n\geq N$}.$$
By the linearity and monotonicity of integration and the preceding corollary, for each $n\geq N$,
$$\left|\int_{E}f – \int_{E}f_n\right| = \left|\int_{E}[f – f_n]\right|\leq\int_{E}\left|f-f_n\right|\leq [\epsilon/\operatorname{m}\left(E\right)]\cdot\operatorname{m}\left(E\right) = \epsilon.$$
Therefore $\lim_{n\to\infty}\int_{E}f_n = \int_{E}f$.
My question is: How come if we switch bounded to uniformly bounded, and uniform convergent to pointwise convergent the proof no longer holds? Moreover, how come we can not longer say: Let $\epsilon>0$. Choose an index $N$ for which
$$\left|f-f_n\right|<\epsilon/\operatorname{m}\left(E\right)\text{ on $E$ for all $n\geq N$}.$$
Is this statement no longer true if $\{f_n\}\to f \text{ pointwise on $E$}$?
Best Answer
If all that you assume is that $(f_n)_{n\in\Bbb N}$ converges pointwise to $f$, then all you know is that $\lim_{n\to\infty}f_n(x)=f(x)$ for each individual $x\in E$. But you cannot deduce from this that, given $\varepsilon>0$, you have $\bigl|f(x)-f_n(x)\bigr|$ for every sufficiently large $N$ and for all $x\in E$.
For instance, take $E=[0,1]$. For each $x\in E$ and each $n\in\Bbb N$, defined$$f_n(x)=\begin{cases}n^2x&\text{ if }x<\frac1{2n}\\-n^2\left(x-\frac1n\right)&\text{ if }x\in\left[\frac1{2n},\frac1n\right]\\0&\text{ if }x>\frac1n,\end{cases}$$Then $(f_n)_{n\in\Bbb N}$ converges pointwise to the null function, but, for each $n\in\Bbb N$, $\int_0^1f_n(x)\,\mathrm dx=\frac14$.