Counter-example to the extended dominated convergence theorem

Question

To add some context, let $(\Omega,\mathcal{F},\mu)$ be a measure space. The statement of the Extended Dominated Convergence Theorem(EDCT) is:

$\lbrace f_{n}\rbrace_{n\geq 1}$ be a sequence of functions from $L^{1}(\Omega,\mathcal{F},\mu)$ such that $f_{n}\to f$ a.e.($\mu$) and there exists a sequence $\lbrace g_{n}\rbrace_{n\geq 1}\in L^{1}(\Omega,\mathcal{F},\mu)$ such that $g_{n}\to g$ a.e.($\mu$), $|f_{n}|\leq g_{n}$ a.e.($\mu$) for all $n\geq 1$ and $\int g_{n}d\mu\to\int gd\mu$. Then, (i) $f\in L^{1}(\Omega,\mathcal{F},\mu)$ (ii) $f_{n}\xrightarrow{L^{1}} f$ (iii) $\int f_{n}d\mu \to \int fd\mu$

I'm stuck while trying to find a counterexample to the above theorem such that, $\lbrace f_{n}\rbrace_{n\geq 1}, f\in L^{1}(\Omega,\mathcal{F},\mu), f_{n}\to f$ a.e.($\mu$) and $\int f_{n}d\mu\to\int f d\mu$ but $||f_{n}-f||_{L^{1}(\Omega,\mathcal{F},\mu)}\not\to 0$.

If the sequence of functions is non-negative, we can see $|f_{n}-f|\leq f_{n}+f=g_{n}$. Also, $\int g_{n}d\mu=\int f_{n}d\mu+\int fd\mu\to2\int fd\mu$. So applying the EDCT we cannot find a counter-example if the the sequence of functions is non-negative.

So is there any such example that shows that a sequence $g_{n}$ that dominates $f_{n}$ is necessary for convergence in $L^{1}$-norm. Thanks in advance for any help.

Answer 1

Counter-example: Consider the measure space $(\mathbb{R},\mathcal{B}(\mathbb{R}),m)$, where $m$ is the usual Lebesgue measure. Let $f_{n}:\mathbb{R}\rightarrow\mathbb{R}$ be defined by $f_{n}(x)=n^{2}1_{(\frac{1}{2n},\frac{1}{n})}(x)-n^{2}1_{(-\frac{1}{n},-\frac{1}{2n})}(x)$. Define $f:\mathbb{R}\rightarrow\mathbb{R}$ by $f(x)=0$. Clearly $f_{n}\rightarrow f$ pointwisely. Moreover $\int f_{n}dm=\int dm=0$. However $||f_{n}-f||_{1}\not\rightarrow0$.

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However, we have:

Let $(X,\mathcal{F},\mu)$ be a measure space. Let $(f_{n})$ be a sequence of real-valued integrable functions defined on $X$. Let $f:X\rightarrow\mathbb{R}$ be an integrable function. Suppose that $f_{n}\rightarrow f$ a.e., then $\lim_{n}\left(||f_{n}||-||f||-||f_{n}-f||\right)=0$, where $||f||=\int|f|d\mu$ etc...

We go to prove the above assertion. Observe that $|f_{n}-f|\leq|f_{n}|+|f|$, so $|f_{n}|-|f|-|f_{n}-f|\geq-2|f|.$ On the other hand, $|f_{n}|\leq|f_{n}-f|+|f|$ and hence $|f_{n}|-|f|-|f_{n}-f|\leq0\leq2|f|$. Combining, we have $\Biggl||f_{n}|-|f|-|f_{n}-f|\Biggr|\leq2|f|$. Clearly, $|f_{n}|-|f|-|f_{n}-f|\rightarrow0$ a.e.. By Dominated Convergence Theorem, we have that $\int\left(|f_{n}|-|f|-|f_{n}-f|\right)d\mu\rightarrow 0$. That is, $||f_{n}||-||f||-||f_{n}-f||\rightarrow0$.

Remark:

1. In particular, if $||f_n||\rightarrow ||f||$, then $||f_n-f||\rightarrow 0$

2. Counter-example does not exist if $f_n$ and $f$ are non-negative.

Answer 2

It's not clear to me whether you're saying EDCT is actually true; on the one hand your question says you're looking for a counterexample, but then the "such that" makes it unclear whether you're looking for a counterexample to EDCT.

In any case, EDCT (as stated above!) is clearly false. It has the following corollary:

Assume EDCT. Suppose $\lbrace f_{n}\rbrace_{n\geq 1}$ is a sequence of functions from $L^{1}(\Omega,\mathcal{F},\mu)$ such that $f_{n}\to f$ a.e.($\mu$). Then, (i) $f\in L^{1}(\Omega,\mathcal{F},\mu)$ (ii) $f_{n}\xrightarrow{L^{1}} f$ (iii) $\int f_{n}d\mu \to \int fd\mu$.

You must know counterexamples to that. But

Proof. Let $g_n=|f_n|$, $g=|f|$. Then $g_n\in L^1$, $g_n\ge|f_n|$ and $g_n\to g$ almost everywhere. So EDCT implies (i), (ii) and (iii).

Note that if you add the hypothesis $\int g_n\to\int g\ge0$ then EDCT is a standard result (also true).