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Let $(\Omega, \mathcal A,\mu)$ be a measure space and $f_k\rightarrow f$ a.e., $f_k\geq0$ and $\int f d\mu\leq C$ for some $C>0$.

How can you show $\int f_k d\mu\leq C$ ?

My attempt: I thought about using Fatou: $$ C\geq\int fd\mu\geq\liminf\int f_kd\mu $$

But this way I only know the limes inferior is $\leq C$.

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If we are using Lebesgue measure on the unit interval, $[0,1]$, we can demonstrate a quick counter example.

Suppose that we have the sequence $f_k(x) = 1+1/k$ where $x\in[0,1]$. This converges a.e. (well everywhere) to the constant function $f(x)=1$ for all $x \in [0,1]$.

Then $\int f_k d\mu =1+1/k$ but $\int f d\mu = 1$. Thus for all $k$ we have $\int f_k d\mu > \int f d\mu = 1$.

Thus we see that $\int f d\mu = 1 = C > 0$, however for all $k$ we have $\int f_k d\mu > C$. Thus showing $\int f d\mu \le C$ is not sufficient for demonstrating $\int f_k d\mu < C$ for sufficiently large $k$.

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    $\begingroup$ How is this a counterexample? Couldn't I say that C = 1, which bounds both integrals in this case? $\endgroup$
    – Gregory
    Jan 20 '15 at 17:45
  • $\begingroup$ The question needs to be more specific here. Are we asking if this can hold for all $k$? Or for sufficiently large $k$? As stated this is a counter example. $\endgroup$
    – Joel
    Jan 20 '15 at 17:47
  • $\begingroup$ The idea is that for any $C > 0$ and larger than $\int f d\mu$ that the inequality holds for $\int f_k d\mu$. You would need to check more than $C=1$ but also $C=1/2$, $C=1/4$ etc. $\endgroup$
    – Joel
    Jan 20 '15 at 17:48
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    $\begingroup$ I think the way to think about this is that $C$ is unknown to $f_k$ beforehand. All we know is that $C$ bounds the integral of the limiting function, and then we look back at the integrals of the sequence and ask if they are under $C$ as well. At least that is how I interpret it. $\endgroup$
    – Joel
    Jan 20 '15 at 17:53
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    $\begingroup$ The forward direction is certainly true by Fatou. However, the backwards direction does not hold generally. $\endgroup$
    – Joel
    Jan 20 '15 at 17:56
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I think this is a counterexample.

Let $f_k=\chi_{[k,\infty)}\to f=0$ everywhere. and use Lebesgue measure.

Btw, Fatou lemma states $\int \liminf f_k\le \liminf \int f_k$

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