# Let $f_n$ be real & measurable w/ $f_n(x)\rightarrow 0$. Show there is a positive measurable function $h$ such that $f_nh\rightarrow 0$ in measure.

Question: Let $$f_n:\mathbb{R}\rightarrow\mathbb{R}$$ be measurable functions such that $$f_n(x)\rightarrow 0$$ as $$n\rightarrow\infty$$ for every $$x\in\mathbb{R}$$. Show there exists a positive measurable function $$h$$ such that $$f_nh\rightarrow 0$$ in measure.

My thoughts: I had two ideas for how to approach this problem: First, I could find some function $$h$$, in terms of $$f_n$$, that converges to $$0$$ pointwise almost everywhere which would then imply convergence to $$0$$ in measure. Or, second, use function $$h$$ (again, I feel like I need it in terms of $$f_n$$) and show that $$f_nh$$ converges to $$0$$ in $$L_2$$ (by integrating), thus in $$L_1$$, and so in measure. However, I am struggling to find such an $$h$$. Maybe it would be best to just try and do it directly by trying to show $$\lim_{n\rightarrow\infty}m(x\in\mathbb{R}:|f_n(x)h(x)|\geq\epsilon)=0$$ for all $$\epsilon>0$$.
Any help would be greatly appreciated! Thank you

It seems that we can get a stronger result that there exists a positive measurable function $$h$$ such that there is a measurable set $$A \subseteq \mathbb{R}$$ such that $$m(\mathbb{R} \setminus A) = 0$$ and $$f_n h \to 0$$ uniformly on $$A$$. Note that this clearly implies that $$f_n h \to 0$$ in measure.

Denote by $$m$$ the standard Lebesgue measure on $$\mathbb{R}$$ and take any expression $$\mathbb{R} = \cup_{i=1}^{\infty} E_i$$ where each $$E_i$$ is measurable and $$m(E_i) < \infty$$ (for example by bounded intervals $$E_i$$'s). For each $$j \geq 1$$, we have, by the basic standard theorem of Egorov (https://en.wikipedia.org/wiki/Egorov%27s_theorem), that there exists a measurable subset $$A_{ij} \subseteq E_i$$ with $$m(E_i \setminus A_{ij}) < 1/j$$ and $$f_n \to 0$$ uniformly on $$A_{ij}$$. Then note that $$m(E_i \setminus \cup_{j=1}^{\infty} A_{ij}) = 0$$. Taking $$A = \cup_{i,j \geq 1} A_{ij}$$, we have also that $$\displaystyle \mathbb{R} \setminus A = (\bigcup_i E_i )\setminus A \subseteq \bigcup_{i \geq 1} (E_i \setminus \bigcup_{j \geq 1} A_{ij})$$, so that $$m(\mathbb{R} \setminus A) = 0$$. Since $$\{ A_{ij} \}_{i,j \geq 1}$$ is a countable family of sets, we can reindex this so that $$A = \cup_{i=1}^{\infty} A_i$$ where each $$A_i$$ is measurable and $$f_n \to 0$$ uniformly on each $$A_i$$.

Set $$B_k = \{ x \, | \, \sup_n |f_n(x)| \leq k \}$$. Since $$f_n \to 0$$ pointwise on $$\mathbb{R}$$, for each $$x \in \mathbb{R}$$ we must have $$\sup_n |f_n(x)| < \infty$$ (why?), so that $$\mathbb{R} = \cup_{k=1}^{\infty} B_k$$.

Now take $$\displaystyle h = \sum_{i,j \geq 1} \frac{1}{j2^{ij}} \chi_{A_i \cap B_j} + \chi_{\mathbb{R} \setminus A}$$, where as usual $$\chi_E$$ denotes the indicator function on $$E$$, i.e. $$1$$ on $$E$$ and $$0$$ elsewhere. Note that $$\displaystyle \sum_{i,j \geq 1} \frac{1}{j 2^{ij}} \leq \sum_{i,j} \frac{1}{2^{ij}} = \sum_i \sum_j \frac{1}{2^{ij}} = \sum_i \frac{1}{2^i - 1} \leq \sum_i \frac{1}{2^{i-1}} = 2 < \infty$$ and that $$h > 0$$ everywhere on $$\mathbb{R}$$, so $$h$$ is a positive (real-valued) measurable function on $$\mathbb{R}$$.

Given $$\epsilon > 0$$, we can take some large positive integer $$s$$ so that $$\displaystyle \frac{4}{2^s - 1} < \frac{\epsilon}{2}$$. With this choice, we have

$$\sum_{i \geq 1} \sum_{j \geq s} \frac{1}{2^{ij}} = \sum_{i=1}^{\infty} \frac{1}{2^{is} - 2^{is - i}} = \sum_{i \geq 1} \frac{1}{2^{is - i}(2^i - 1)} \leq \sum_{i \geq 1} \frac{1}{2^{is -i}(2^{i-1})} = \sum_{i \geq 1} \frac{1}{2^{is - 1}} = 2 \sum_{i \geq 1} \frac{1}{2^{is}} = \frac{2}{2^s - 1} < \frac{\epsilon}{4}.$$

Now,

$$\displaystyle |f_n h| \leq \sum_{1 \leq i,j \leq s} \frac{|f_n|}{j 2^{ij}} \chi_{A_i \cap B_j} + \sum_{i \geq s, j \geq 1} \frac{|f_n|}{j 2^{ij}} \chi_{A_i \cap B_j} + \sum_{j \geq s, i \geq } \frac{|f_n|}{j 2^{ij}} \chi_{A_i \cap B_j} + |f_n| \chi_{\mathbb{R} \setminus A} \leq \sum_{1 \leq i,j \leq s} \frac{|f_n|}{j 2^{ij}} \chi_{A_i \cap B_j} + \sum_{i \geq s, j \geq 1} \frac{1}{2^{ij}} \chi_{A_i \cap B_j} + \sum_{j \geq s, i \geq 1} \frac{1}{2^{ij}} \chi_{A_i \cap B_j} + |f_n| \chi_{\mathbb{R} \setminus A}.$$

In the very latter expression, the middle two terms sum to something smaller than $$\frac{\epsilon}{2}$$, and in the first term, $$\sum_{1 \leq i,j \leq s} \frac{|f_n|}{j 2^{ij}} \chi_{A_i \cap B_j}$$, we have uniform convergence to $$0$$ (it's a finite sum of sequences of functions that converge uniformly to $$0$$), so there exists $$N$$ such that if $$n \geq N$$ then $$\sum_{1 \leq i,j \leq s} \frac{|f_n|}{j 2^{ij}} \chi_{A_i \cap B_j} < \epsilon / 2$$. So that for $$n \geq N$$, we have $$|f_n(x) h(x)| < \epsilon$$ for all $$x \in A$$. So $$f_n h \to 0$$ uniformly on $$A$$.

• Thank you so much for the elaborate answer! I was wondering, even though this rersult is stronger, would there be a nice (concise) way to do the original problem? Jul 1, 2021 at 22:02