Can anyone help me to prove the following:
If $f$ and $g$ are in $\mathfrak{L}(\mu )$ on $E$, then prove the following: (i) If $a\leq f(x)\leq b$ for $x\in E$, and if $\mu \left(E\right)<\infty $, then $a\mu(E) \le \int_E f d\mu \le b \mu(E) $ (ii) If $f(x)\le g(x)$ for $x\in E$, then $\int_E f d\mu \le \int_E g d\mu$.
I have referred to "Walter Rudin Principles of Mathematical analysis Page number 315".
For $(ii)$ I have the following proof: Let $f(x) \leq g(x)$ for $x \in E$. Then $f^+ = \text{max}(f,0) \leq \text{max}(g, 0) = g^+$ and $f^- = -\text{min}(f, 0) \geq -\text{min}(g, 0) = g^-$ (by definition) ($\because~ \text{min}(f, 0) \leq \text{min}(g, 0)$).
''$\therefore~ \int\limits_E f^+ d \mu \leq \int\limits_E g^+ d\mu$ and $\int\limits_E f^- d \mu \geq \int\limits_Eg^- d\mu$ (by definition)''
\textbf{The above step is not cleared to me}
Hence $\int\limits_E f d \mu = \int\limits_E f^+ d\mu - \int\limits_E f^- d\mu \leq \int\limits_E g^+ d\mu - \int\limits_E g^- d \mu = \int\limits_E g d\mu$
Thus $\int\limits_E f d \mu \leq \int\limits_E g d\mu.$
Please provide me elaborated proofs for $(i)$ and $(ii)$. Thank you
I am not entirely sure what I am not allowed to use here...
If $f,g$ are integrable then so is $h=g-f $
If $h \ge 0$ on $E$, then $\int_E h \ge 0$.
Since $\int_E g = \int_E f + \int_E h$ we see that $ \int_E g \ge \int_E f$.
The first result follows noting that $\int_E a d \mu = a \mu(E)$ and similarly for $b$.