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I am not sure if the question fits MO standards as it is an elementary measure theory question (and if it's not, expect to be tazered by the MO police). Here goes an answer, anyway.

To expand on Robin Chapman's comment, first, the theorem as stated is false withouth the assumption that $E$ has finite measure. The correct generalization is the Lebesgue dominated convergence where the sequence $f_{n}$ is such that there is an integrable $g$ such that $\|f_n{x}\|\leq g$.

To see why, it fails without the boundedness condition consider the sequence of intervals $E_n= [0, 1/n]$ and take the sequence $(n\chi(E_n))$ where $\chi(E_n)$ is the characteristic function (or indicator functions) of $E_n$. This sequence converges pointwise to $0$ but

$\int n\chi(E_n) = 1$


so that the sequence of integrals does not converge to $0$. What is happening is that you are shrinking the support of the functions but the same time increasing their "amplitude" so that the two cancel each other out and the integral stays constant while the functions themselves converge to zero. The uniform bound on the sequence, prevents their "amplitudes" of running off to infinity and screwing up the integrals.

Note: examples

Examples can be concocted where the convergence is uniform instead of just pointwise. The idea is to do the reverse of the previous example: shrink the amplitude of the functions (to guarantee their uniform convergence) while enlarging their support. This will need a measure space of infinite measure. I will leave that as an exercise.

Regards, G. Rodrigues

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I am not sure if the question fits MO standards as it is an elementary measure theory question (and if it's not, expect to be tazered by the MO police). Here goes an answer, anyway.

To expand on Robin Chapman's comment, first, the theorem as stated is false withouth the assumption that $E$ has finite measure. The correct generalization is the Lebesgue dominated convergence where the sequence $f_{n}$ is such that there is an integrable $g$ such that $\|f_n{x}\|\leq g$.

To see why, it fails without the boundedness condition consider the sequence of intervals $E_n= [0, 1/n]$ and take the sequence $(n\chi(E_n))$ where $\chi(E_n)$ is the characteristic function (or indicator functions) of $E_n$. This sequence converges pointwise to $0$ but

$\int n\chi(E_n) = 1$


so that the sequence of integrals does not converge to $0$. What is happening is that you are shrinking the support of the functions but the same time increasing their "amplitude" so that the two cancel each other out and the integral stays constant while the functions themselves converge to zero. The uniform bound on the sequence, prevents their "amplitudes" of running off to infinity and screwing up the integrals.

Note: examples can be concocted where the convergence is uniform instead of just pointwise. I will leave that as an exercise.

Regards, G. Rodrigues