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I have read a conclusion in a textbook:

Suppose $f_n,f$ are density functions of some r.v. also $f_n\to f$ a.e., then $$\int f_n \mathrm{d}x \to\int f \mathrm{d}x $$

Fisrt I want to use "DCT".since$$\int |f_n -f |\mathrm{d}x\ \leq2$$ even though I can't find a dominated function.

But I soon found the "DCT" is false. A counterexample is $$n\chi_{\{0\leq x\leq \frac{1}{n}\}}$$

Now the question is how to construct the dominated function or avoid using the DCT?

Lookout
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  • Maybe you are trying to use $|f_n-f| \to 0$ and $|f_n - f| \le |f_n|+|f|\le 2|f|$ and hence more generally: $|f_n-f|^p \to 0$ and $|f_n - f|^p \le 2^p |f|^p$ where $p\ge 1$. – Squirtle Aug 23 '14 at 04:54
  • Sorry this^ is just a useful corollary. Namely, I've shown if LDCT hypothesis holds then $\lim_n \int_X |f_n-f|d\mu=0$ because $\lim_n \int_X |f_n-f|d\mu=\int_X 0 d\mu = 0$. – Squirtle Aug 23 '14 at 05:03
  • This seems very relevant to your question:http://math.stackexchange.com/questions/149619/does-bounded-covergence-theorem-hold-for-riemann-integral – Squirtle Aug 23 '14 at 05:05
  • The statement $\int f_n \mathrm{d}x \to\int f \mathrm{d}x$ is a bit vague: do you mean the convergence of CDFs? I edited the title assuming this; if I'm wrong, you can revert or edit again. –  Aug 23 '14 at 05:18
  • If $f_n$ and $f$ are density functions, then $\int f_n = \int f = 1$ for all $n$, so $\int f_n \to \int f$ trivially. I think you want to prove something different than what you wrote. – Will Nelson Aug 23 '14 at 09:19

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It is perhaps easiest to directly apply Fatou's Lemma : $$ 2 = \int \liminf (f_n+f - |f_n-f|) \leq \liminf \int(f_n+f - |f_n-f|) $$ $$ = 2 - \limsup \int|f_n -f| $$ and so $$ \lim \int |f_n-f| = 0 $$ which proves what you want.

Prahlad Vaidyanathan
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HINT:

Assuming $|f_n|\le g$ a.e., $f_n \to f$ a.e., and $g\ge 0 $ is integrable; use the fact that $g+f_n$ is nonnegative and converges to $g+f$ and and apply Fatou's Lemma to conclude

$$\int_X f d\mu \le \lim \inf_n \int_X f_n d\mu$$

Now repeat this argument with $g-f_n$ instead to show that

$$\lim \sup_n \int_X f_n d\mu \le \int_X f d\mu$$

Squirtle
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