1. ## Monotone Decreasing

Let $\displaystyle (f_n)$ be a non-negative sequence of monotone decreasing integrable functions on $\displaystyle [0,1]$.
Show that $\displaystyle (f_n) \downarrow$ 0 if and only if $\displaystyle (\int_{[0,1]} f_n) \downarrow$ 0.

2. Originally Posted by problem
Let $\displaystyle (f_n)$ be a non-negative sequence of monotone decreasing integrable functions on $\displaystyle [0,1]$.
Show that $\displaystyle (f_n) \downarrow$ 0 if and only if $\displaystyle (\int_{[0,1]} f_n) \downarrow$ 0.
Since the functions are non-increasing (monotome decreasing) it means that $\displaystyle f_n(b) \leq f_n(x) \leq f_n(a)$ for each $\displaystyle x\in [a,b]$. Therefore, $\displaystyle f_n(b) \leq \smallint_0^1 f_n \leq f_n(a)$. However, we are told that $\displaystyle f_n(a),f_n(b)\to 0$ therefore the integral, being squeezed between those two sequences, must go to zero too. Conversely it is not true. Just let $\displaystyle f_n(a) = 1$ with $\displaystyle f_n(x) = 0$ for $\displaystyle a<x\leq b$. Then each $\displaystyle f_n(x)$ is integrable, non-negative, non-increasing with zero integral however $\displaystyle f_n$ does not converge to the zero function.

3. the monotone decreasing that i refer to is that $\displaystyle 0\le f_{n+1} \le f_n$ for all n.Is it still valid to use the argument above?