Exercise 7.4.1

Question

Let $f$ be a bounded function on a set $A$, and set
$$
\begin{gathered}
M=\sup \{f(x): x \in A\}, \quad m=\inf \{f(x): x \in A\}, \
M^{\prime}=\sup \{|f(x)|: x \in A\}, \quad 	ext { and } \quad m^{\prime}=\inf \{|f(x)|: x \in A\} .
\end{gathered}
$$
\begin{enumerate}[label=(\alph*)] 
\item Show that $M-m \geq M^{\prime}-m^{\prime}$.
\item Show that if $f$ is integrable on the interval $[a, b]$, then $|f|$ is also integrable on this interval.
\item Provide the details for the argument that in this case we have $\left|\int_{a}^{b} fight| \leq \int_{a}^{b}|f|$.
 \end{enumerate}

Ulisse Mini · Accepted Answer

\begin{enumerate}[label=(\alph*)] 
\item If $f(x) \geq 0$ then $M' = M, m' = m$, and the result is trivial. Similarly if $f(x) \leq 0$ then $M' = -m,\ m' = -M$. Finally if $f(x)$ has both positive and negative values then $M \geq 0,\ m \leq 0,\ M' = \max\{M, -m\},\ m' \geq 0,\ M - m = M + (-m) \geq M' \geq M' - m'$
\item Given part (a), if we have a partition $P$ so that $U(f, P) - L(f,P) < \epsilon$, then we also have $U(|f|, P) - L(|f|, P) \leq U(f, P) - L(f,P) \epsilon$.
\item $f \leq |f|$, so $\int^b_a f \leq \int^b_a |f|$. Similarly $\int^b_a f \geq \int^b_a -|f| = -\int^b_a |f|$, which together means $\left|\int_a^b f\right| \leq \int_a^b|f|$
 \end{enumerate}

Exercise 7.4.1

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Exercise 7.4.1

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