# Real-valued measurable functions $f_n \geq f_{n+1}$ with $f_n \to 0$ pointwise, then $f_n \to 0$ in measure

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.

• $f_n$ 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.) – 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.