Vitali's Convergence Theorem but one hypothesis changes

Question

We have the following problem:

Let $(Y, \Gamma , \nu)$ be a measure space. Suppose that $\{g_{n}\} \, \subset \, L^{p} \, := \, L^{p}(Y,\Gamma , \nu).$ Prove that $\lim_{n} g_{n} = g$ in $L^p$ if and only if:

1) For each $\epsilon > 0$ $$\lim_{n\to\infty} \nu(\{y \in Y \, : \, |g_{n}(y)-g(y)| > \epsilon \}) = 0.$$ 2) If $\epsilon > 0$ exists $\delta(\epsilon) > 0$ such that if $E \in \Gamma$ and $\nu(E) < \delta(\epsilon)$, then $$\left|\int_{E} g \, d\nu \right| < \epsilon . $$
3) For each $\epsilon > 0$ exists $N \in \mathbb{N}$ and a set $E_{\epsilon} \in \Gamma$ with $\nu(E_{\epsilon}) < \infty$ such that $$ \int_{E_{\epsilon}^{c}} \left| g_{n} \right| ^{p} \, d\nu < \epsilon \quad \forall n \geq N. $$

As you can see the conditions are pretty similar to the Vitali's Convergence Theorem, except for the second one.

I have been following the proof in the book of Rober G. Bartle called Element of Integration but I can't change that proof so that fits in my problem.

So far the idea that I had is consider $\epsilon>0$ and let $E_{\epsilon}$ be as in 3). Then apply the Minkowski inequality to the function $g_n-g=(g_n-g)\chi_{E_{\epsilon}}+g_n\chi_{E_{\epsilon}^c}+(-g)\chi_{E_{\epsilon}^c}$.

From there I can narrow the second and the first elements, but not the last one. I would really appreciate some help.

Thank you so much.

Answer 1

The statement is wrong: Take $g_n = n 1_{(0,1/n]}$. Then $g_n(x) \rightarrow 0$ for all $x \in \mathbb{R}$. The measure space under consideration should be $(\mathbb{R},\mathcal{B}(\mathbb{R}),\lambda)$, i.e. the Lebesgue measure on $\mathbb{R}$. Then we have $\lambda(|g_n|> 0) = 1/n \rightarrow 0$. Since $g=0$, the condition (2) is always satisfied. In (3) you can take $E_\varepsilon = [0,\varepsilon]$. Then $\nu(E_\varepsilon) < \varepsilon$ and for all $n \ge N$ with $1/N < \varepsilon$ we have $$\int_{E_\varepsilon^c} |g_n| d \lambda =0,$$ i.e. the modifacted version of Kavi Rama Murthy holds.

The problem is that (2) should be replaced by the condition that for all measurable sets $A$ with $\nu(A)< \delta$ $$\int_{E} |g_n| \, d \nu < \varepsilon \quad \text{for all } n \in \mathbb{N}.$$ By taking $E \cap \{g_n \ge 0 \}$, resp. $E \cap \{g_n \le 0 \}$ the last conidition can be replaced by $$\left| \int_{E} g_n \, d \nu \right| < \varepsilon \quad \text{for all } n \in \mathbb{N}.$$ This is exactly the condition which can be found in the corresponding Wikipedia article, as cited in the comments. Note that we don't need to make the modification as made in Kavi Rama Murthy answer.

Answer 2

As I mentioned in one of the comments 3) should say $\nu (E_{\epsilon}) <\epsilon$. Here is how you handle the last term in your decomposition: by going to a subsequence we may replace 1) by almost everywhere convergence. Let $\epsilon >0$ Then we have $\int_{E_{\epsilon}^{c}} |g_n|^{p}d\nu <\epsilon$ for $n$ sufficiently large. By Fatou's Lemma this gives $\int_{E_{\epsilon}^{c}} |g|^{p}d\nu \leq\delta$. Hence $\|-gI_{E_{\epsilon}^{c}}\|_p \leq \epsilon^{1/p}$.