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Problem Statement

Suppose $(X,\mathcal{A},\mu)$ is a measure space, each $f_n$ is integrable and non-negative, $f_n \rightarrow f$ a.e, and $\int f_n \rightarrow \int f$. prove that for each $A \in \mathcal{A}$,$$\int_A f_n \rightarrow \int_A f$$

Background

My text just introduced Monotone Convergence Theorem, Fatou's Lemma, and Lebesgue Dominated Convergence Theorem.

Attempt

  • Theorem: Suppose $f_n$, $g_n$, $f$, and $g$ are integrable, $f_n \rightarrow f$ a.e, $g_n \rightarrow g$ a.e, $|f_n|\leq g_n$ for each $n$, and $\int g_n \rightarrow \int g$. Then $\int f_n \rightarrow \int f$.

    Proof: Since $\int_N h \ d\mu=0$ on a null set $N$, we may redefine $f_n$, $g_n$, $f$ and $g$ to be $0$ on the appropriate nullsets and consider the convergence on the entire space.$^{(1)}$ Next, since $|f_n|\leq g_n$, we have $g+f>0$ and $g-f>0$. We use Fatou's lemma to furnish the following two arguments. First, \begin{align*} \int g+f \ d\mu&=\int\lim_{n \to \infty}\inf \left(f_n+g_n\right) \ d\mu \\ &\leq \lim_{n \to \infty}\inf \int f_n+g_n \ d\mu\\ &=\lim_{n \to \infty}\inf \int f_n\ d\mu+\int g \ d\mu, \end{align*} the last line following from the fact that $\int g_n \rightarrow \int g$. Thus, $$\int_f \leq \lim_{n \to \infty}\inf \int f_n\ d\mu.$$ similarly, \begin{align*} \int g-f \ d\mu &=\int\lim_{n \to \infty}\inf \left(g_n-f_n\right) \ d\mu \\ &\leq \lim_{n \to \infty}\inf \int g_n-f_n \ d\mu, \end{align*} so that \begin{eqnarray*} - \int f \ d\mu &\leq \lim_{n \to \infty}\inf \int-f_n \ d\mu\\ &=\lim_{n \to \infty}\sup \int-f_n \ d\mu. \end{eqnarray*} Thus $$\int f \geq \lim_{n \to \infty}\sup \int f_n \ d\mu,$$ and we have the result.

  • Solution:

    Note that in our problem, $f_n\chi_A \rightarrow f\chi_A$, $f \rightarrow f_n$, $|f_n\chi_A|\leq f_n$ for each $n$, and $\int f_n \rightarrow \int f$. Thus, our theorem yields $\int_A f_n \rightarrow \int_A f$. (Note that $\chi$ is the characteristic function$.

My Question(s)

Is this correct? I felt awkward with $(1)$ but I think it was the right way to deal with the a.e convergence.

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  • $\begingroup$ I think your statement (1) is fine. It almost doesn't seem to be necessary. At what point did you actually need it? $\endgroup$ – Andrew Jun 17 '15 at 3:16
  • $\begingroup$ Since the problem mentioned a.e convergence, I felt that I had to deal with it in some dramatic way somehow..otherwise I didn't do anything too different from the standard proof of Fatou's lemma. (I should mention that the theorem I have listed was the previous problem to the problem being asked). $\endgroup$ – illysial Jun 17 '15 at 3:17
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    $\begingroup$ I think it's okay. If you're learning measure theory for the first time it doesn't hurt to include these kinds of statements. As you get more comfortable with the subject you might come to a point where you don't write these statements anymore, but they aren't wrong. $\endgroup$ – Andrew Jun 17 '15 at 3:23
  • $\begingroup$ See also this question: math.stackexchange.com/q/678282 $\endgroup$ – saz Jun 17 '15 at 5:29

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