The following is an exercise from Carothers' Real Analysis:

Suppose $f$ and $f_n$ are nonnegative, measurable functions, that $f=\lim_{n\to\infty} f_n$ and that $\int f=\lim_{n\to\infty}\int f_n<\infty$. Prove that $\int_E f=\lim_{n\to\infty}\int_E f_n$ for any measurable set $E$. (Hint: Consider both $\int_E f$ and $\int_{E^c} f$.) Give an example showing that this need not be true if $\int f=\lim_{n\to\infty}\int f_n=\infty$.


Using Fatou's Lemma, I can get that

$$\int_E f=\int_E \lim_{n\to\infty} f_n=\int_E\liminf_{n\to\infty} f_n\leq \liminf_{n\to\infty}\int_E f_n=\lim_{n\to\infty}\int_E f_n$$

However, I'm doubtful that this is correct since the hint in the book said to consider both $E^c$ and $E$. I know that if $E$ is measurable, then its complement is measurable as well. A small push in the right direction would be appreciated. Thanks.


Consider the following:

$$\int f = \int_E f + \int_{E^c} f \le \varliminf \int_E f_n + \varliminf \int_{E^c} f_n \le \varliminf\left(\int_E f_n + \int_{E^c} f_n\right) = \varliminf \int f_n = \int f$$

  • $\begingroup$ I'm afraid I'm still having some trouble concluding. Could you please elaborate a little more on your answer? I don't see why showing that $\int f\leq \int f$ helps us here. Thanks. $\endgroup$ – Sujaan Kunalan Mar 30 '15 at 16:21
  • $\begingroup$ You're missing that the inequalities imply $\varliminf \int_E f_n + \varliminf \int_{E^c} f_n = \int_E f + \int_{E^c} f$. What can you infer from this? $\endgroup$ – kobe Mar 30 '15 at 16:23
  • $\begingroup$ If I'm not mistaken, I believe it means I can infer that $\liminf_{n\to\infty} \int_E f_n=\int_E\lim_{n\to\infty} f_n=\int_E f$. $\endgroup$ – Sujaan Kunalan Mar 30 '15 at 16:49
  • $\begingroup$ And since we know the integral is finite, the limit exists and hence must equal the liminf. So we can conclude that $\liminf_{n\to\infty} \int_E f_n=\lim_{n\to\infty}\int_E f=\int_E\lim_{n\to\infty} f_n=\int_E f$? $\endgroup$ – Sujaan Kunalan Mar 30 '15 at 17:06
  • 2
    $\begingroup$ No, the point was $\liminf \int_{E^c} f = \int_{E^c} f$ implies $\color{red}{\limsup} \int_E f_n = \int_E f$. So since $\liminf \int_E f_n = \int_E f$ and $\limsup \int_E f_n = \int_E f$, we have $\lim \int_E f_n = \int_E f$. $\endgroup$ – kobe Mar 30 '15 at 19:24

Call $g_n=f_n \chi_E $, $g=f\chi_E$, then we have $g_n\leq f_n$ so that applying Fatou's lemma to $f_n-g_n$ we arrive at $$ \int f-g \leq\liminf_n \int f_n-g_n= \lim_n\int f_n- \limsup_n \int g_n= \int f-\limsup_n \int g_n, $$ by hypothesis this is equivalent to $$ \limsup_n \int g_n \leq \int g. $$ Combine this with the inequality you got to conclude.


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