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.

  • $\begingroup$ $f_n$ 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

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.


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.


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