兩題分析 (Analysis Quiz #8)

1. Let $\{f_k\}$ be a sequence of Lebesgue integrable functions such that $f_k \rightarrow f$ a.e. with $f$ also integrable. Then $\int |f_k - f| \rightarrow 0 $ iff $\int |f_k|\rightarrow \int |f|$.

2. Let $f$ be Lebesgue integrable over a measurable set $E$. Then for any $\epsilon > 0$, there is a simple function $\phi$ such that $\int_E |f-\phi| < \epsilon$.

Solution. 

1. 
($\Rightarrow$) By Fatou's lemma, 
\[\int \liminf |f_k|\leq \liminf\int|f_k|.\] 
Therefore, we have 
\[\int |f|\leq \liminf\int|f_k|.\]
One the other hand, since $|f-f_k|+|f|\geq |f_k|$,
\[\limsup\int |f_k| \leq \limsup \int |f_k-f| + \int |f|\]
for all $k$. Because $\limsup \int|f_k-f| \rightarrow 0$ (by assumption), we conclude that 
\[\int |f| \leq \liminf \int |f_k| \leq \limsup \int|f_k| \leq \int |f|.\]

($\Leftarrow$) Since $|f-f_k| \leq |f|+|f_k|$, 
\[\int \liminf (|f|+|f_k|-|f-f_k|) \leq \liminf \int |f|+|f_k|-|f_k-f|.\]
So,
\[2\int |f| \leq 2\int |f|-\int |f_k-f|.\]
Because $f_k\rightarrow f$ a.e. , We may deduce that $\int |f_k-f| \leq 0$. 
Therefore,  $\int |f_k-f|=0$ since the integral is nonnegative.

2.
Note that there exist simple functions $g_k\nearrow f^+$ and $h_k\nearrow f^-$. 
By 1, given $\epsilon > 0$, there exists $N>0$ such that 
\[\int |f^+-g_N| < \epsilon /2 \]
and
 \[\int |f^--h_N|< \epsilon /2.\]
Therefore, let $\phi = g_N-h_N$, 
\[\int|f-\phi|\leq \int |f^+-g_N|+\int |f^--h_N|< \epsilon/2+\epsilon/2 = \epsilon.\]

小感：第一題的 ($\Leftarrow$) 手法有些技巧，我在小考的時候想破頭想不到怎麼做。