0
$\begingroup$

Let $(X, \mathcal{A}, \mu)$ be a measure space with $\mu$ finite measure. Let $f_n : (X, \mathcal{A}) \to (\mathbb{R}, \mathcal{B}(\mathbb{R}))$ be measurable for every $n \in \mathbb{Z}_+$ such that $f_n \to 0$ as $n \to \infty$ pointwise and $f_n \geq f_{n+1}$ for every $n$ pointwise. Does this imply that $f_n \to 0$ in measure?

I think this is false, however, I cannot find a counterexample. The reason I think this is false is that the functions $f_n$ may not be non-negative, therefore we cannot, for instance, let $g_n:= -f_n$ and apply the Monotone Convergence Theorem and the Markov inequality to conclude.

$\endgroup$
  • $\begingroup$ $f_n$ is necessarily non-negative because $f_n\geq\lim_{N\to\infty}f_N=0$ pointwise. (If the pointwise convergence is replaced by '$\mu$-almost everywhere pointwise convergence', the conclusion is still true in $\mu$-almost everywhere sense.) $\endgroup$ – Sangchul Lee Sep 27 '19 at 14:11
2
$\begingroup$

This is true. Suppose that $f_n$ does not converge to $0$ in measure. Then there is an $\varepsilon>0$, $\delta > 0$ and a strictly increasing sequence of naturals $n_k$ such that $$\mu(|f_{n_k}| > \varepsilon) > \delta$$ for every $k$.

To conclude, let $$A = \bigcap_{k \in \mathbb{N}} \{ |f_{n_k}| > \varepsilon \}.$$ Notice that $\{|f_{n_{k+1}}| > \varepsilon\} \subseteq \{|f_{n_{k}}| > \varepsilon\}$ so that, since $\mu$ is finite we have that $$\mu(A) = \lim_{k \to \infty} \mu(|f_{n_k}| > \varepsilon) \geq \delta > 0.$$

But then $f_{n_k} \not \to 0$ pointwise on $A$ which gives a contradiction.

$\endgroup$
1
$\begingroup$

Under the mentioned conditions ($f_{n}\to0$ pointwise and $f_{n}\geq f_{n+1}$ for every $n$) the functions must be non-negative.

For some fixed $\epsilon>0$ let $A_{n}:=\left\{ x\in X\mid\left|f_{n}\left(x\right)\right|<\epsilon\right\} =\left\{ x\in X\mid f_{n}\left(x\right)<\epsilon\right\} $.

Then $A_{1}\subseteq A_{2}\subseteq A_{3}\subseteq\cdots$ and $\bigcup_{n=1}^{\infty}A_{n}=X$.

Then $\mu A_{n}\uparrow\mu X$ so that $\mu A_{n}^{\complement}=\mu X-\mu A_{n}\downarrow0$.

This for every $\epsilon>0$ so we have:$$\forall\epsilon>0\left[\lim_{n\to\infty}\mu\left(\left\{ x\in X\mid\left|f_{n}\left(x\right)\right|\geq\epsilon\right\} \right)=0\right]$$

Conclusion: $f_{n}$ converges in measure to the zero function.

Essential is here that measure $\mu$ is finite.

$\endgroup$

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.