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$f_{n}$ be a sequence of function in $L^{p}[0,1](1<p<\infty)$ and $f_{n}\to f$ pointwise a.e and $f\in L^{p}[0,1]$, and suppose that there is a constant $M$, such that $||f_{n}||_{p} \leq M$ for all $n$. Then for each function $g$ in $L^{q}[0,1]$, we have $$\int fg = \lim \int f_{n}.g $$

I know this problem has been posted a numerous times but I did a proof by myself and I want it to be verified. Most proofs I suppose use the Egoroff's theorem but I here used the Vitali's convergence theorem which is as follows:

If $E$ be a set of finite measure and $f_{n}$ be a sequence of function on $E$ which is uniformly integrable. If $f_{n}\to f$ a.e on $E$, then $f$ is integrable and $$\lim_{n\to \infty}\int_{E} f_{n} = \int_{E}f$$

Now we have $f_{n}.g \to f.g$ a.e on $[0,1]$. I just want to prove that the sequence $f_{n}.g$ is uniformly integrable. So given $\epsilon >0$, I have to produce a $\delta >0$ such that for every $A \subseteq [0,1]$ with $m(A)< \delta$ then $\int_{A} |f_{n}.g| < \epsilon$. Now since $|g|^{q}$ is already integrable we have for given $\epsilon >0$ , then there exists a $\delta >0$ such that for all $A\subseteq [0,1]$ with $m(A)< \delta$ we have $$\int_{A} |g|^{q}<\left(\frac{\epsilon}{M}\right)^{q}.$$ Now $$\int_{A}|f_{n}.g| \leq ||f_{n}||_{p}\left(\int_{A} |g|\right)^{\frac 1q}\leq M.\left(\left(\frac{\epsilon}{M}\right)^{q}\right)^{1/q}< \epsilon$$ when $m(A)<\delta$.

Hence the sequence $f_{n}.g$ is uniformly integrable and the result holds by using Vitali Convergence theorem.

Thanks in advance!

Davide Giraudo
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Riju
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  • Not sure if this will help: https://math.stackexchange.com/questions/1180248/if-f-n-to-f-a-e-and-are-bounded-in-lp-norm-then-int-f-n-g-to-int-fg-f – Wolfy Jul 21 '17 at 23:42

1 Answers1

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In short: yes, it is correct.

In his answer, Did showed that condition (C) below is equivalent to (C1) and (C2) together.

(C) For every $\varepsilon\gt0$, there exists a finite $c$ such that, for every $X$ in $\mathcal H$, $\mathrm E(|X|:|X|\geqslant c)\leqslant\varepsilon$.

(C1) There exists a finite $C$ such that, for every $X$ in $\mathcal H$, $\mathrm E(|X|)\leqslant C$.

(C2) For every $\varepsilon\gt0$ there exists $\delta\gt0$ such that, for every measurable $A$ such that $\mathrm P(A)\leqslant\delta$ and every $X$ in $\mathcal H$, $\mathrm E(|X|:A)\leqslant\varepsilon$.

As discussed in this thread, the condition you checked (namely, (C2)) is not enough in general to guarantee uniform integrability.

However, in the context of the unit interval endowed with Lebesgue measure, (C2) implies (C1) (using the interval $(j\delta,(j+1)\delta)$).

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Davide Giraudo
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  • I didn't understand the discussion in the thread that you gave But are you saying that the there something more to show that $f_{n}.g$ is uniformly integrable? – Riju Jul 11 '17 at 20:29
  • Ths the definiton of uniform integrability in Royden: A family $F$ of measurable function on $E$ is said to be uniformly integrable over $E$ provided for each $\epsilon >0$, there is a $\delta >0$ such that for each $f\in F$ if $A\subset E$ is such that $m(A)<\delta$, then $\int_{A}|f| < \epsilon$ – Riju Jul 11 '17 at 20:30
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    I forgot the fat that you worked with the unit interval endowed with the Lebesgue measure. In this case, what you did is sufficient to guarantee the uniform integrability. I believed you were using the "official" definition of uniform integrability. – Davide Giraudo Jul 11 '17 at 20:37