# Proving two properties of the Riemann integral

I'm working on a proof following theorem (from T.Tao's Analysis 1 book):

Let $I$ be a bounded interval and $f,g:I\to\mathbb{R}$ Riemann integrable functions on $I$; then:

a) $f+g$ is Riemann integrable and $\int_I (f+g)=\int_I f +\int_I g$;

b) $cf$ is Riemann integrable and $\int_I (cf)=c\int_I f\ \forall c\in\mathbb{R}$;

c) $f-g$ is Riemann integrable and $\int_I (f-g)=\int_I f -\int_I g$;

d) $f(x)\geq 0\forall x\in I\Rightarrow \int_I f\geq 0$;

e) $f(x)\geq g(x) \forall x\in I\Rightarrow \int_I f\geq \int_I g$;

f) $f(x):=c \forall x\in I\Rightarrow \int_I f=c|I|$;

g) $J$ bounded interval containing $I$ ($J\subseteq I$), $F\colon J\to\mathbb{R}$ such that $F(x):=f(x)$ if $x\in I$, $F(x):=0$ if $x\notin I\Rightarrow F$ is Riemann integrable on $J$ and $\int_J F=\int_I f$;

h) $\{J,K\}$ partition of $I$ into two intervals $J$ and $K\Rightarrow f_{|J}\colon J\to\mathbb{R},f_{|K}\colon K\to\mathbb{R}$ are integrable on $J$ and $K$ respectively and $\int_I f=\int_J f_{|J}+\int_K f_{|K}$.

Now, I managed to prove a) to f) but I'm struggling with properties g) and h) (in both cases I tried to shows that the upper and lower sums of the functions considered could be written as upper and lower sums of f, but my approach failed to consider that these sums change when we change the domain since the sums are done over the partitions and the partitions may change when we change domain).

So, I'd appreciate any hint about how to prove these last two properties.

Note: the definitions used are as follows:

$f$ is Riemann integrable on interval $I$ if $\overline{\int_I}f=\underline{\int_I}f$ where $\overline{\int_I}f=\inf\{U(f,P):P$ partition of $I\}$, $\underline{\int_I}f:=\sup\{L(f,P):P$ partition of $I\}$ with $U(f,P):=\sum_{J\in P:J\neq\emptyset} (\sup_{x\in J}f(x))|J|$ and $L(f,P):=\sum_{J\in P: J\neq\emptyset}(\inf_{x\in J} f(x))|J|$ and a partition $P$ of a bounded interval $I$ being a finite set of bounded intervals contained in $I$ such that every $x\in I$ lies in exactly one of the bounded intervals $J$ in $P$.

For part g), let $*$ denote the condition that $P$ is a partition of $I$ such that $J$ is a union of intervals in $P$ (I just don't want to write that long condition in formulas). Then $$\overline{\int_I} f \leq \inf\{U(f,P)\mid *\}, \quad \underline{\int_I}f \geq \sup\{L(f,P)\mid *\}.$$ Then $$\sup\{L(f,P) \mid *\} \leq \underline{\int_I}f = \overline{\int_I}f \leq \inf\{U(f,P) \mid *\}.$$ Use this to conclude $\sup\{L(f,P) \mid *\}= \inf\{U(f,P) \mid *\}$. Now note that $L(f,P)$ where $P$ satisfies $*$ is precisely the same as $L(f,P_1)+L(f,P_2)$ where $P_1$ is a partition of $J$ and $P_2$ is a partition of $I\setminus J$. Same with the upper sums. So (justification left up to you), we get $$\sup\{L(f,P_1)\}+\sup\{L(f,P_2)\}=\inf\{U(f,P_1)\}+\inf\{U(f,P_2)\}$$ (again $P_1$ is a partition of $J$ and $P_2$ a partition of $I\setminus J$). You also know the sup of the lower sums is $\leq$ the inf of the upper sums, so this is an equation of the form $A+B =A'+B'$, $A \leq A'$, $B \leq B'$. Conclude that $A=A'$. Finally argue that $A=\underline{\int_I}F$ and $A'=\overline{\int_I}F$.
For part h), note that $f_{|J}$ and $f_{|K}$ are integrable by part g). Also, any partition of $J$ and $K$ together yield a partition of $I$, so $\overline{\int_J}f_{|J}+ \overline{\int_K}f_{|K} \geq \overline{\int_I}f$, and $\underline{\int_J}f_{|J}+\overline{\int_K}f_{|K} \leq \underline{\int_I}f$. Now use the fact that $f$, $f_{|J}$ and $f_{|K}$ are all integrable (on $I$, $J$, and $K$, respectively).
• I thought about using g) to prove h) but to do that one has to view the restriction $f_{|J}\colon J\subseteq I\to\mathbb{R}$ defined as $f_{|J}(x):=f(x)$ if $x\in J$ and $f_{|J}(x):=0$ if $x\notin J$ but in the text $f_{|J}$ is defined as $f_{|J}(x):= f(x)$ if $x\in J$ and $f_{|J}(x)$ is left undefined when $x\notin J$. – lorenzo May 16 '17 at 16:35
• @lorenzo If you are integrating $f_{|J}$ over the interval $J$, then it does not matter whether it is defined outside of $J$ or not. I.e., if $g(x)=f(x)$ on $J$ and $g$ is undefined outside of $J$, and in particular $\int_J f_{|J}=\int_J g$. – kccu May 16 '17 at 16:38