Exercise 1.8 (Basic properties of the real-valued integral)

Question

Let $f, g: \Omega \rightarrow {\bf R}$ be real-valued absolutely integrable functions, and let $c \in {\bf R}$.
\begin{enumerate}
	\item (Linearity) Show that $f+g$ and $cf$ are also real-valued absolutely integrable functions, with
		$$\displaystyle \int_\Omega f+g\ d\mu = \int_\Omega f\ d\mu + \int_\Omega g\ d\mu$$
    and
    $$\displaystyle \int_\Omega cf\ d\mu = c \int_\Omega f\ d\mu.$$
    (For the second relation, one may wish to first treat the special cases $c>0$ and $c=-1$.)
	\item Show that if $f$ and $g$ are equal almost everywhere, then
    $$\displaystyle \int_\Omega f\ d\mu = \int_\Omega g\ d\mu.$$
	\item  Show that $\int_\Omega |f|\ d\mu \geq 0$, with equality if and only if $f$ is zero almost everywhere.
	\item (Monotonicity) If $f \leq g$ almost everywhere, show that $\int_\Omega f\ d\mu \leq\int_\Omega g\ d\mu$.
	\item (Markov inequality) Show that $\mu( \{ \omega: |f(\omega)| \geq t \} ) \leq \frac{1}{t} \int_\Omega |f|\ d\mu$ for any $0 < t < \infty$.
\end{enumerate}

Elu Thingol · Accepted Answer

$
ewcommand{\intleb}[1]{\int_{\Omega} #1 \,\mathrm{d}\mu}$
\begin{enumerate}
	\item 	extit{Linearity}
ewline
	Imagine for a second that we have proven the following identity:
			\begin{align*}
			&(f + g)_+ - (f+g)_- = (f_+ - f_-) + (g_+ - g_-)
			\end{align*}
			Using this, we would have managed to represent the identities only using the operation of addition $+$ as follows (this is necessary to make use of the definition of the signed integral).
			$$(f + g)_+ + f_- + g_- =  (f+g)_- + f_+ + g_+$$
			In other words, both sides would then be represented by unsigned simple functions, and so, using the linearity of the unsigned integral from \citew{exercise_1-3-1} we would be able to arrive at the following step:
			$$\intleb{(f+g)_+} + \intleb{f_-} + \intleb{g_-} = \intleb{(f+g)_-} + \intleb{f_+} + \intleb{g_+} .$$
			Simple arithmetic manipulations would then lead us to
			$$\intleb{(f+g)_+} - \intleb{(f+g)_-} =  \intleb{f_+} - \intleb{f_-} + \intleb{g_+} - \intleb{g_-} .$$
			And thus,
			$$\intleb{(f+g)}=\intleb{f} + \intleb{g}$$
			as desired.
ewline
			It is thus only left to verify that the identity we have used
			$$\forall x \in \mathbb{R}^d: \quad (f + g)_+(x) - (f+g)_-(x) \stackrel{!}{=} (f_+(x) - f_-(x)) + (g_+(x) - g_-(x))$$
			or in other words,
			$$\max\{f(x)+g(x),0\} - \max\{-f(x)-g(x),0\} \stackrel{!}{=} \max\{f(x),0\} - \max\{-f(x),0\} + \max\{g(x),0\} - \max\{-g(x),0\}$$
			is indeed valid. To do so, tediously verify the possible cases for an arbitrary $x\in\mathbb{R}^d$:
			\begin{enumerate}
				\item[1)] $f(x) + g(x) \geq 0$. From this we already have $\max\{f(x)+g(x),0\}-\max\{-f(x)-g(x),0\} =$ $\max\{f(x)+g(x),0\}$. This induces further cases
					\begin{enumerate}
						\item[1.1)] $f(x) \geq 0$ and $g(x) \leq 0$.
						\item[1.2)] $f(x) \geq 0$ and $g(x) > 0$.
						\item[1.3)] $f(x) < 0$ and $g(x) \geq 0$.
						\item[1.4)] $f(x) <0$ and $g(x) < 0$ is impossible.
					\end{enumerate}
				\item[2)] $f(x) - g(x) < 0$.
			\end{enumerate}
			(in all cases try to transform $\max\{f(x)+g(x),0\}-\max\{-f(x)-g(x),0\}$.)
ewline

We now demonstrate the scalar multiplication property. We consider several cases:      
			\begin{itemize}
				\item $c\geq 0$
ewline
					Notice that for any real number $y$ we then have $\max\{cy,0\} = c\max\{y,0\}$, and thus
					\begin{align*}
					\intleb{cf} 
					&= \intleb{(cf)_+} - \intleb{(cf)_-} = \intleb{c(f)_+} - \intleb{c(f)_-}\
					&= c \cdot \intleb{f_+} - c\cdot \intleb{f_-} = c \cdot \intleb{f}
					\end{align*}
				\item $c = -1$
ewline
					We then have $(-f)_+ = \max\{-f,0\} = f_-$ and similarly $(-f)_- = f_+$; thus,        
					\begin{align*}
					\intleb{-f} &= \intleb{(-f)_+} - \intleb{(-f)_-} = \intleb{f_-} - \intleb{f_+}\ &= -1 \cdot \left(\intleb{f_+} - \cdot \intleb{f_-}ight) = -\intleb{f}
					\end{align*}
				\item $c < 0$
ewline
					Follows by combining the first and the second case.
			\end{itemize}

\item 	extit{Equivalence}
ewline
	Notice that $f \stackrel{a.e.}{=} g$ trivially implies that $f_+ \stackrel{a.e.}{=} g_+$ and $f_- \stackrel{a.e.}{=} g_-$ (for instance by $\{x\in\mathbb{R}^d: f_+(x) 
eq g_+(x)\} \subseteq \{x\in\mathbb{R}^d: f(x) 
eq g(x)\}$). The equality of the intervals then follows directly from \citew{exercise_1-5} by applying it to both positive and negative part of the real valued simple integral.
	
	\item	We have
	$$\int_\Omega |f| \,\mathrm{d}\mu = \int_\Omega |f|_+ \,\mathrm{d}\mu - \int_\Omega |f|_- \,\mathrm{d}\mu = \int_\Omega |f|_+ \,\mathrm{d}\mu \geq 0.$$
	If $|f|\stackrel{a.e.}{=} 0$, then by the previous part of the exercise we must also have $\int |f| = \int 0 = 0$. If $\int |f| = \int |f|_+ = 0$ then $|f|_+ \stackrel{a.e.}{=}0$ by \citew{exercise_1-5}. 
	
	\item 	extit{Monotonicity}
ewline
	Since $g-f \stackrel{a.e.}{\geq} 0$, we have $\int (g-f) = \int |g-f|$. By the previous part of the exercise, we obtain
	$$\int_\Omega g \,\mathrm{d}\mu - \int_\Omega f \,\mathrm{d}\mu = \int_\Omega g-f \,\mathrm{d}\mu = \int_\Omega |g-f| \,\mathrm{d}\mu \geq 0.$$

\item 	extit{Markov inequality}
ewline
	We have the trivial pointwise inequality
	$$\lambda 1_{\{x\in\Omega: f(x)\geq\lambda\}} \leq f(x).$$
	We apply integral from both sides and using the compatibility of Lebesgue measure with the Lebesgue integral (\href{./exercise_1-5}{Exercise 1.5}), we conclude
	$$\lambda m(\{x\in\Omega: f(x)\geq\lambda\}) \leq \intleb{f}.$$
	
\end{enumerate}

Exercise 1.8 (Basic properties of the real-valued integral)

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Exercise 1.8 (Basic properties of the real-valued integral)

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