Some nice properties of Lesbesgue Integrals
Lesbesgue's Integral is very well behaved. This is shown by Theorem 1.27 and Fatou's Lemma. Proofs are a result of application of Monotone Convergence theorem.
Theorem 1.27
If $f_n:X \rightarrow [0,\infty]$ is measurable, for $n=1,2,3,\cdots$ and
\begin{equation}
f(x)=\sum_{i=1}^\infty f_n(x) \text{ } ( x \in X)
\end{equation}
then
\begin{equation}
\int_X f d\mu = \sum_{i=1}^\infty \int_x f_n d\mu
\end{equation}
In otherwords, the Lebesgue Integrals respect sums!
Proof: Start off with two functions $f_1$ and $f_2$. Know that each of these functions have simple measurable functions $s_i'$ and $s_i^{ii}$ such that $s_i^{'}\rightarrow f_1$ and $s_i^{''} \rightarrow f_2$. Let $s=s_1+s_2$. Then clearly $s \rightarrow f_1+f_2$. Since we have a sequence of simple functions whose limits are functions $f_1,f_2$, we can apply monotone convergence theorem. In case of $s_i^{'}$ we have, \begin{equation} \int s_i^{'} d\mu = \int f_1 d\mu \end{equation} Similarly, \begin{equation} \int s_i^{''} d\mu = \int f_2 d\mu \end{equation} Integrals of simple functions can be expanded. Hence, \begin{equation} \int s_i d\mu = \int (s_i^{'}+s_i^{''})d\mu = \int s_i^{'}d\mu + \int s_i^{''} d\mu \end{equation} Again applying Monotone convergence theorem, we have \begin{equation} \int (f_1+f_2)d\mu = \int f_1 d\mu + \int f_2 d\mu \end{equation} Now let $g_N = f_1+f_2+\cdots+f_N$. Clear $g_N \rightarrow f$. Apply induction on the last equation, we get \begin{equation} \int g_N d\mu = \sum_{i=1}^N f_i d\mu \end{equation} Again apply Monotone convergence theorem to LHS of above, \begin{equation} \int f d\mu = \sum_{i=1}^\infty f_i d\mu \end{equation} Fatou's Lemma: If $f_n \rightarrow [0,\infty]$ is measurable for each positive integer $n$, then \begin{equation} \int_X \left( lim_{n\rightarrow \infty}\, inf f_nd\mu \right) \leq lim_{n\rightarrow \infty}\, inf \int_Xf_n d\mu \end{equation} Proof: Put \begin{equation} g_k(x) = inf f_k(x) \text{ for } k=1,2,\cdots \end{equation} Clear that $g_k(x) \leq f_k(x)$. Then, \begin{equation} \int_Xg_k(x)d\mu \leq \int_x f_k(x) d\mu \text{ for } k=1,2,\cdots \end{equation} Also, $0\leq g_1(x) \leq g_2(x) \cdots$. It follows each $g_k$ is measurable. Then, \begin{equation} g_k(x) \rightarrow lim_{k \rightarrow \infty} \text{inf} f_k(x) \end{equation} Using monotone convergence theorem, \begin{equation} \int_Xg(k) d\mu \rightarrow \int_X \text{lim}_{n \rightarrow \infty} \text{ inf} f_n(x) d\mu \end{equation} Conclusion follows.
Proof: Start off with two functions $f_1$ and $f_2$. Know that each of these functions have simple measurable functions $s_i'$ and $s_i^{ii}$ such that $s_i^{'}\rightarrow f_1$ and $s_i^{''} \rightarrow f_2$. Let $s=s_1+s_2$. Then clearly $s \rightarrow f_1+f_2$. Since we have a sequence of simple functions whose limits are functions $f_1,f_2$, we can apply monotone convergence theorem. In case of $s_i^{'}$ we have, \begin{equation} \int s_i^{'} d\mu = \int f_1 d\mu \end{equation} Similarly, \begin{equation} \int s_i^{''} d\mu = \int f_2 d\mu \end{equation} Integrals of simple functions can be expanded. Hence, \begin{equation} \int s_i d\mu = \int (s_i^{'}+s_i^{''})d\mu = \int s_i^{'}d\mu + \int s_i^{''} d\mu \end{equation} Again applying Monotone convergence theorem, we have \begin{equation} \int (f_1+f_2)d\mu = \int f_1 d\mu + \int f_2 d\mu \end{equation} Now let $g_N = f_1+f_2+\cdots+f_N$. Clear $g_N \rightarrow f$. Apply induction on the last equation, we get \begin{equation} \int g_N d\mu = \sum_{i=1}^N f_i d\mu \end{equation} Again apply Monotone convergence theorem to LHS of above, \begin{equation} \int f d\mu = \sum_{i=1}^\infty f_i d\mu \end{equation} Fatou's Lemma: If $f_n \rightarrow [0,\infty]$ is measurable for each positive integer $n$, then \begin{equation} \int_X \left( lim_{n\rightarrow \infty}\, inf f_nd\mu \right) \leq lim_{n\rightarrow \infty}\, inf \int_Xf_n d\mu \end{equation} Proof: Put \begin{equation} g_k(x) = inf f_k(x) \text{ for } k=1,2,\cdots \end{equation} Clear that $g_k(x) \leq f_k(x)$. Then, \begin{equation} \int_Xg_k(x)d\mu \leq \int_x f_k(x) d\mu \text{ for } k=1,2,\cdots \end{equation} Also, $0\leq g_1(x) \leq g_2(x) \cdots$. It follows each $g_k$ is measurable. Then, \begin{equation} g_k(x) \rightarrow lim_{k \rightarrow \infty} \text{inf} f_k(x) \end{equation} Using monotone convergence theorem, \begin{equation} \int_Xg(k) d\mu \rightarrow \int_X \text{lim}_{n \rightarrow \infty} \text{ inf} f_n(x) d\mu \end{equation} Conclusion follows.
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