# Show that $f_n1_{A_n}$ convergences in mean

Consider the measurable space $(\Omega,\mathcal{A},\mu)$. Let $f,f_1,f_2,\ldots$ be measurable functions on that measurable space and $A,A_1,A_2,\ldots\in\mathcal{A}$. Let $(f_n)$ converge in mean to $f$ and $1_{A_n}$ to $1_A$ in measure. Show that then $f_n1_{A_n}$ converges in mean to $f1_A$.

Hello and good evening!

My idea is to show instead that $(f_n1_{A_n})_{n\in\mathbb{N}}$ converges to $f1_A$ in measure and that $(f_n1_{A_n})_{n\in\mathbb{N}}$ is uniformly integrable (because this together is equivalent to the convergence in mean of $(f_n1_{A_n})_{n\in\mathbb{N}}$ to $f1_A$).

I. Uniformly integrability

Consider any $\varepsilon>0$. $(f_n)_{n\in\mathbb{N}}$ is uniformly integrable (and converges to $f$ in measure what is needed below). So there exists an integrable function $h\geq 0$, so that $$\sup_{f_n\in (f_n)_{n\in\mathbb{N}}}\int 1_{\lvert f_n\rvert\geq h}\lvert f_n\rvert\, d\mu<\varepsilon.$$ For this function $h$, it is $$(\lvert 1_{A_n}f_n\rvert-h)^+\leq 1_{\lvert 1_{A_n}f_n\rvert\geq h}\lvert 1_{A_n}f_n\rvert\leq 1_{\lvert f_n\rvert\geq h}\lvert f_n\rvert.$$ So it follows that $$\int (\lvert 1_{A_n}f_n\rvert-h)^+\, d\mu\leq\int 1_{\lvert f_n\rvert\geq h}\lvert f_n\rvert\, d\mu\leq\sup_{f_n\in (f_n)_{n\in\mathbb{N}}}\int 1_{\lvert f_n\rvert\geq h}\lvert f_n\rvert\, d\mu<\varepsilon$$ what means that $$\sup_{1_{A_n}f_n\in (1_{A_n}f_n)_{n\in\mathbb{N}}}\int (\lvert 1_{A_n}f_n\rvert-h)^+\, d\mu<\varepsilon.$$ So $(1_{A_n}f_n)_{n\in\mathbb{N}}$ is uniformly integrable.

II. Convergence in measure

Consider any $\varepsilon>0$. It is $$\left\{\omega\in\Omega:\lvert 1_{A_n}f_n-1_Af\rvert>\varepsilon\right\}\\=\left\{\omega\in\Omega:\lvert 1_{A_n}f_n-1_{A_n}f+1_{A_n}f-1_Af\rvert>\varepsilon\right\}\\\subset \left\{\omega\in\Omega:\lvert 1_{A_n}f_n-1_{A_n}f\rvert+\lvert 1_{A_n}f-1_Af\rvert>\varepsilon\right\}\\\subset\left\{\omega\in\Omega:\lvert 1_{A_n}f_n-1_{A_n}f\rvert>\frac{\varepsilon}{2}\right\}\cup\left\{\omega\in\Omega:\lvert 1_{A_n}f-1_Af\rvert>\frac{\varepsilon}{2}\right\}\\=\left\{\omega\in\Omega:\lvert 1_{A_n}(f_n-f)\rvert>\frac{\varepsilon}{2}\right\}\cup\left\{\omega\in\Omega:\lvert f(1_{A_n}-1_A)\rvert>\frac{\varepsilon}{2}\right\}\\\subset\left\{\omega\in\Omega:\lvert f_n-f\rvert>\frac{\varepsilon}{2}\right\}\cup\underbrace{\left\{\omega\in\Omega:\lvert 1_{A_n}-1_A\rvert=1\right\}}_{=\left\{\omega\in\Omega:\lvert 1_{A_n}-1_A\rvert>\lambda\right\}\text{ for a }0<\lambda<1}.$$ So it is $$\mu(\left\{\omega\in\Omega:\lvert 1_{A_n}f_n-1_Af\rvert>\varepsilon\right\})\leq\mu(\left\{\omega\in\Omega:\lvert f_n-f\rvert>\frac{\varepsilon}{2}\right\}\cup\left\{\omega\in\Omega:\lvert 1_{A_n}-1_A\rvert>\lambda\right\})\\\leq\underbrace{\mu(\left\{\omega\in\Omega:\lvert f_n-f\rvert>\frac{\varepsilon}{2}\right\})}_{\to 0}+\underbrace{\mu(\left\{\omega\in\Omega:\lvert 1_{A_n}-1_A\rvert>\lambda\right\})}_{\to 0}\to 0$$ $$\Longrightarrow \mu(\left\{\omega\in\Omega:\lvert 1_{A_n}f_n-1_Af\rvert>\varepsilon\right\})\to 0$$

From I. and II. it follows that $(1_{A_n}f_n)_{n\in\mathbb{N}}$ converges to $1_Af$ in mean.

Could you pls say me if my proof is allright?

With best wishes for X-mas,

math12

• part 2 is correct. In fact just helped someone else to do this... Dec 25, 2013 at 20:32
• Dont quite follow part 1. What is h? Dec 25, 2013 at 20:35
• Why so you assume fn is uniformly integrable? Dec 25, 2013 at 20:45
• Yes, I saw that you helped in another thread to clarify the convergence in measure. It inspired me, so thank you for that!
– user34632
Dec 25, 2013 at 23:17
• You need to tell me, are you proving this for finite measures or infinite measures? Also i assume $\int f_n <\infty$? Dec 26, 2013 at 1:14

Does uniform integrability plus convergence in measure imply convergence in $L^1$?
• Your definition there is, that $\sup_f\int_{\lvert f\rvert > h}\lvert f\rvert\, d\mu$ goes to $0$ if $h\to\infty$, right? Our definition is for any $\varepsilon>0$ there is a nonnegative, integrable function $h$ so that $\sup_f\int_{\lvert f\rvert\geq h}\lvert f\rvert\, d\mu<\varepsilon$. Is that really the same?