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Let $\{f_n\}$ be a sequence of Lebesgue integrable functions and $g: [0,\infty)\to [0,\infty)$ be an increasing and continuous function such that $\displaystyle\lim_{x\to\infty}g(x) = \infty$.

We also have that

(i) $\int_0^1|f_n(x)|g(|f_n(x)|)\,dx < 100$

(ii) $ f_n \to 0$ almost everywhere

We must show that

$$ \lim_{n\to\infty}\int_0^1 |f_n(x)| \, dx = 0$$

What I have:

Let $\epsilon > 0$,

$g$ is continuous and increasing, then

$$ \exists M > 0, g(|f_n(x)|) > M \mbox{ and } \frac{100}{M} < \frac{\epsilon}{2}$$

Now consider the following partition of $[0,1]$

$$E_1 = \{x \in [0,1]: |f_n(x)| \leq \frac{1}{M}\}$$ $$E_2 = \{x \in [0,1]: |f_n(x)| > \frac{1}{M}\}$$

Then $$\int_0^1 |f_n(x)|\,dx = \int_{E_1}|f_n(x)| + \int_{E_2}|f_n(x)| \, dx$$

Thus,

\begin{align} & \left|\int_0^1|f_n(x)|\,dx\right| < \frac{1}{M}m(E_1) + \int_0^1 |f_n(x)| \frac{g(|f_n(x)|)}{g(|f_n(x)|)} \, dx \\[10pt] < {} & \frac{1}{M} + \frac{1}{M} \int_0^1 |f_n(x)|g(|f_n(x)|) \, dx < \frac{100}{M} + \frac{100}{M} < \frac{\epsilon}{2} + \frac{\epsilon}{2} = \epsilon. \end{align}

I want to know if I got the proof right. Thanks.

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  • $\begingroup$ You lost me at the line "$g$ is continuous and increasing, then [...]." Can you explain what you argue there? $\endgroup$ – Clement C. Aug 19 '17 at 0:10
  • $\begingroup$ $g$ is a fixed function: the fact that it's an increasing function doesn't mean that "it goes to infinity at every point" (such statement does not even mean anything), which I guess is what you seem to say. $\endgroup$ – Clement C. Aug 19 '17 at 0:25
  • $\begingroup$ To clarify, take $g(x)=x$ in the question, and check your proof with this specific instance. (This $g$ is indeed strictly increasing, so satisfies the assumption.) $\endgroup$ – Clement C. Aug 19 '17 at 0:26
  • $\begingroup$ @RichardClare: I feel confused about your question now. What if we assume $g$ is constant 0? That means the first condition doesn't exist. But obviously, condition 2 is not enough for the conclusion. I guess $g$ should be strictly increasing. $\endgroup$ – Selene Aug 19 '17 at 3:26
  • $\begingroup$ @XIAODAQU In English, the convention is that "increasing" means "strictly increasing" (otherwise, it is "non-decreasing"). It makes little sense, but anyway... $\endgroup$ – Clement C. Aug 19 '17 at 3:34
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$\forall \varepsilon>0$

$\exists M>0,\ s.t.\ 100/g(M)<\varepsilon/3$.

Due to Egoroff. $\exists E(measurable)\subset[0, 1]\ \&\ m([0, 1]-E)<\varepsilon/(3M)$. And $f_n\xrightarrow{u.}0$ on E.

Find an enough large N such that $\forall n>N$, $|f_n|<\varepsilon/3$ on E.

Then we have

\begin{eqnarray} \int_0^1|f_n|dx&=&\int_E|f_n|dx+\int_{[0, 1]-E}|f_n|dx\\ &\leq&\varepsilon/3+\int_{([0, 1]-E)\cap\{|f_n|\leq M\}}|f_n|dx+\int_{([0, 1]-E)\cap\{|f_n|> M\}}|f_n|dx\\ &\leq&\varepsilon/3+\int_{[0, 1]-E}Mdx+\int_{[0, 1]\cap\{|f_n|>M\}}|f_n|dx\\ &\leq&2\varepsilon/3+\int_{[0, 1]\cap\{|f_n|>M\}}|f_n|\frac{g(|f_n|)}{g(|f_n|)}dx\\ &\leq&2\varepsilon/3+\frac{1}{g(M)}\int_0^1|f_n|g(|f_n|)dx\\ &\leq&\varepsilon \end{eqnarray}

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  • $\begingroup$ Now everything makes sense to me. Thank you!! $\endgroup$ – Richard Clare Aug 19 '17 at 17:37
  • $\begingroup$ Hey, I want to ask you something. The Egoroff theorem says that $f_n \to 0$ uniformly on $[0,1]\setminus E$. $|f_n| < \frac{\epsilon}{3}$ in $[0,1]\setminus E$ not in E. ... I now have my doubts. $\endgroup$ – Richard Clare Aug 19 '17 at 18:10
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    $\begingroup$ @RichardClare: I don't quite understand what you mean...but I think you've taken my "$[0, 1]-E$" as your "$E$". Maybe in your statement, $f_n$ are uniformly converges in $[0, 1]-E$ and $m(E)<\varepsilon$. I just express them oppositely. It's all the same. $\endgroup$ – Selene Aug 20 '17 at 2:29
  • $\begingroup$ Thank you for the clarification. $\endgroup$ – Richard Clare Aug 20 '17 at 2:33

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