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Let $μ$ a measure on a set $X$ and let $g_{n}∈L_{1}(X, μ)$ a decreasing sequence of integrable functions. Suppose that $g_{n}(x) → g(x)$. prove $$\lim_{n\to\infty}\int g_{n}(x)dμ=0\iff g (x)=0~~~~\text{a.e.} $$

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Hint: Apply Monotone Convergence theorem. –  Stefan Hansen Jan 29 '13 at 7:26
    
but the functions are decreasing and are not necessarily positive. –  Alexander Osorio Jan 29 '13 at 7:34
    
Hint: apply it to $-g_n$ –  Ilya Jan 29 '13 at 7:37
    
but the functions can be non-negative or non-positive or both.and the monotone convergence theorem is for sequences no-negative. –  Alexander Osorio Jan 29 '13 at 7:48
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Alexander: There are special cases of monotone convergence you may be most familiar with, but it applies in the following situations: (1) The sequence is increasing and bounded below by an integrable function. In this case, $+\infty$ is a possible value of the limit. (2) The sequence is decreasing and bounded above by an integrable function. In this case, $-\infty$ is a possible value of the limit. This is the case that applies to your situation, because all functions are bounded above by $g_1$. To get the version of MCT you like, consider $(g_1-g_n)$. –  Jonas Meyer Jan 29 '13 at 7:50
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up vote 3 down vote accepted

This is false. Consider $g(x)=x$ on $[-1,1]$ with Lebesgue measure, and $g_n=g$ for all $n$.

If you meant to include the hypothesis that $g_n\geq 0$ for all $n$, then as Stefan commented, applying the monotone (or dominated) convergence theorem (or Fatou's lemma) is a good idea.

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+1 Easy example :) –  AD. Jan 29 '13 at 7:54
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