Background Information:

Theorem 2.30 - Suppose that $\{f_n\}$ is Cauchy in measure. Then there is a measurable function $f$ such that $f_n\rightarrow f$ in measure, and there is a subsequence $\{f_{n_j}\}$ that converges to $f$ a.e. Moreover, if also $f_n\rightarrow g$ in measure, then $g=f$ a.e.


Exercise 33 - If $f_n\geq 0$ and $f_n\rightarrow f$ in measure then $\int f\leq \liminf \int f_n$

Attempted proof - Let $f_n\geq 0$ and $f_n\rightarrow f$ in measure then for every $\epsilon >0$ $$\mu\left(\{x:|f_n(x) - f(x)|\geq\epsilon\}\right)\rightarrow 0 \ \ \text{as} \ \ n\rightarrow \infty$$ Thus by Theorem 2.30 there is subsequence $\{f_{n_j}\}$ which converges to $f$ a.e. Then by Fatou's lemma $$\int f \leq \int \liminf f_{n_j} \leq \liminf \int f_{n_j}$$

I am not sure where to go from here any suggestions is greatly appreciated

  • 7
    $\begingroup$ Your question history suggests you are having severe difficulties with the problems in Folland. Have you considered seeking instruction on this material, or if you already are, seeking help from your instructor? $\endgroup$ – Eric Towers Jul 1 '16 at 16:47
  • 2
    $\begingroup$ @EricTowers I am not having "severe difficulties". I am preparing for my Quals. I know I am posting many questions in regards to Folland but this is only because my mentor is helping me and making sure I understand his theorems and some of his problems. $\endgroup$ – Wolfy Jul 1 '16 at 16:50

@Wolfy, The idea in your attempt proof is correct but it fails basically because when we take a subsequence $\{f_{n_j}\}$ of $\{f_n\}$, we have
$$\liminf \int f_n \leq \liminf \int f_{n_j}$$

The "trick" is start by peeking a subsequence $\{f_{n_i}\}_i$ such that $$\int f_{n_i} \to \liminf_{n \to \infty} \int f_n $$ in other words: $$\lim_{i \to \infty}\int f_{n_i} = \liminf_{n \to \infty} \int f_n $$

Here is the proof in details.

Exercise 33 - If $f_n\geq 0$ and $f_n\rightarrow f$ in measure then $\int f\leq \liminf \int f_n$


Since $\{ \int f_n \}_n$ is just a sequence of complex number, there is subsequence $\{f_{n_i}\}_i$ such that $$\int f_{n_i} \to \liminf_{n \to \infty} \int f_n $$

Since $f_n\rightarrow f$ in measure, we have that $f_{n_i}\rightarrow f$ in measure. So there is a subsequence $\{f_{n_{i_j}}\}_j$ of $\{f_{n_i}\}_i$ such that $$f_{n_{i_j}} \to f \textrm{ a.e.} \tag{1} $$

Since $\{f_{n_{i_j}}\}_j$ is a subsequence of $\{f_{n_i}\}_i$, we also have $$\int f_{n_{i_j}} \to \liminf_{n \to \infty} \int f_n \tag{2}$$

So we have from $(1)$ and $(2)$, using Fatou's lemma,

$$ \int f = \int \lim_{j \to \infty } f_{n_{i_j}} = \int \liminf_{j \to \infty } f_{n_{i_j}}\leq \liminf_{j \to \infty } \int f_{n_{i_j}}= \lim_{j \to \infty } \int f_{n_{i_j}}= \liminf_{n \to \infty} \int f_n$$


Just do it in two steps. First pass to a subsequence $f_{n_j}$ so that $$\int f_{n_j} \to \liminf \int f_n$$ then pass to a further subsequence $f_{n_{j_\ell}}$ which converges to $f$ pointwise a.e. (possible since $f_{n_j}$ still converges to $f$ in measure) and then apply Fatou's Lemma.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.