proofs of riemann integral properties

If \[f:\left[ a,b \right]\to\mathbb{R}\] is Riemann integrable and \[c\in\mathbb{R}\], then \[c\cdot f\] is integrable and \[\int_{a}^{b}cf=c\int_{a}^{b}f\].

Suppose \[c\ge0\]. Then, for any set \[A\] that is a subset of \[\left[ a,b \right]\], \[\sup_{A}(c\cdot f)=c\cdot\sup_{a}f\] and \[\inf_{A}(c\cdot f)=c\cdot\inf_{a}f\]. It follows that \[U(c\cdot f;P)=c\cdot U(f;P)\] for every partition \[P\], as \[\sum_{k=1}^{n}[c\cdot (M_{k}\cdot (x_{k}-x_{k-1}))]=c\cdot\sum_{k=1}^{n}[M_{k}\cdot (x_{k}-x_{k-1})]\].

Taking the infimum over the set \[\Pi\] of all partitions of \[\left[ a,b \right]\] (refer back to the Riemann integral article for explanations on symbols), or simply the infimum of all upper Riemann sums calculated with different types of partitions, \[U(cf)=\inf_{P\in\Pi}U(cf;P)=c\cdot\inf_{P\in\Pi}U(f;P)=c\cdot U(f)\]. Similarly, by this logic, for lower Riemann sums, \[L(cf)=\sup_{P\in\Pi}L(cf;P)=c\cdot\sup_{P\in\Pi}L(f;P)=c\cdot L(f)\].

Then if \[f\] is Riemann integrable, it will fulfill the condition \[U(cf)=c\cdot U(f)=c\cdot L(f)=L(cf)\], which implies that \[c\cdot U(f;P)-c\cdot L(f;P)<\epsilon\implies U(f;P)-L(f;P)<\frac{\epsilon}{c}\] (with the exception of \[c=0\], where it is trivially true since \[cf=0\] collapses to zero). Since \[f\] is integrable, then we should be able to find some finite partition with sufficiently small intervals that fulfills the condition \[U(f;P)-L(f;P)<\frac{\epsilon}{c}\].

Thus, this shows that \[\int_{a}^{b}cf=c\int_{a}^{b}f\] when \[c\ge0\].

Now, consider \[-f\]. Since \[\sup_{A}(-f)=-\inf_{A}f\] and \[\inf_{A}(-f)=-\sup_{A}f\] (as we basically flipped the graph upside down), then \[U(-f;P)=-L(f;P)\] and \[L(-f;P)=-U(f;P)\]. This because \[U(f;P)=\sum_{k=1}^{n}M_{k}\cdot \left( x_{k}-x_{k-1} \right)\], where \[M_{k}\] is the supremum of its respective subinterval, then as a consequence of \[\sup_{A}(-f)=-\inf_{A}f\], the upper Riemann sum would be flipped over to become the lower Riemann sum (as the supremum has been flipped over to become the infimum). It works similarly for lower Riemann sums.

Therefore, \[U(-f)=\inf_{P\in\Pi}U(-f;P)=-\sup_{P\in\Pi}L(f;P)=-L(f)\] and \[L(-f)=\sup_{P\in\Pi}L(-f;P)=-\inf_{P\in\Pi}U(f;P)=-U(f)\]. This implies that \[U(-f;P)-L(-f;P)=(-L(f;P))-(-U(f;P))=U(f;P)-L(f;P)<\epsilon\].

Hence, \[-f\] is Riemann integrable if \[f\] is integrable and \[\int_{a}^{b}(-f)=-\int_{a}^{b}f\].

Finally, if \[c<0\], we take \[c=- \left| c \right|\]. As a consequence what we've just shown above, \[\int_{a}^{b}(-\left| c \right|f)=-\left| c \right|\int_{a}^{b}f=c\int_{a}^{b}f\].

If \[f\] and \[g\] are bounded, then \[f+g\] is bounded and \[\sup_{I}(f+g)\le \sup_{I}f+\sup_{I}g\]. This can be trivially proven by: \[f(x)\le\sup f\] and \[g(x)\le\sup g\], hence \[(f+g)(x)=f(x)+g(x)\le\sup f+\sup g\], and it follows that \[\sup(f+g)\le\sup f+\sup g\]. Similarly, \[\inf_{I}(f+g)\ge\inf_{I}f+\inf_{I}g\].

Let \[\omega_{f}(I)\] be the oscillation of \[f\] on an interval \[I\] such that \[\omega_{f}(I)=\sup_{x\in I}f(x)-\inf_{x\in I}f(x)\]. Then \[\omega_{(f+g)}(I)\le\omega_{f}(I)+\omega_{g}(I)\]. In other words, \[\omega_{f}(I)\] measures the distance between the "minimum" and "maximum" of \[f\]-values on the interval \[I\], therefore the distance measured in function \[f+g\] can not be larger than \[f\] and \[g\] added separately.

In general, when we add two different functions together, we can not expect to be able to just add their Riemann sums together. Instead, we prove the equality by estimating the upper and lower integrals of \[f+g\] from above and below by those of \[f\] and \[g\].

If \[f,g:\left[ a,b \right]\to\mathbb{R}\] are Riemann integrable functions, then \[f+g\] is Riemann integrable, and \[\int_{a}^{b}(f+g)=\int_{a}^{b}f+\int_{a}^{b}g\].

We first prove that if \[f,g:\left[ a,b \right]\to\mathbb{R}\] are bounded, but not necessarily Riemann integrable, then \[U(f+g)\le U(f)+U(g)\] and \[L(f+g)\ge L(f)+L(g)\].

Suppose that \[P=\left\{ I_{1},I_{2},\dots,I_{n} \right\}\] is a partition of \[\left[ a,b \right]\]. Then:

\begin{align*} U(f+g;P)&=\sum_{k=1}^{n}\sup_{I_{k}}(f+g)\cdot \left| I_{k} \right|\\ &\le\sum_{k=1}^{n}\sup_{I_{k}}f\cdot \left| I_{k} \right|+\sum_{k=1}^{n}\sup_{I_{k}}g\cdot \left| I_{k} \right|\\ &\le U(f;P)+U(g;P) \end{align*}

Let \[\epsilon>0\]. Since the upper integral, \[U(f)\], is the infimum of all possible upper Riemann sums, \[\inf_{P\in\Pi}U(f;P)\], there must exist partitions \[Q\] and \[R\] such that \[U(f;Q)<U(f)+\frac{\epsilon}{2}\] and \[U(g;R)<U(g)+\frac{\epsilon}{2}\]. In other words, \[U(f)\] is the "smallest" possible upper Riemann sum, so for any \[\epsilon>0\], there exists upper Riemann sum with a less "fine" partition, \[Q\] (as there are infinite ways to partition \[\left[ a,b \right]\]), such that it's smaller than \[U(f)+\frac{\epsilon}{2}\]. The similar applies to \[U(g;R)\].

Then, let \[P\] be a partition that includes all the endpoints in partition \[Q\] and \[R\] (and maybe containing even more additional points), meaning that \[P\] is a more "finely" defined partition (which gives a more accurate upper Riemann sum), it follows that \[U(f;P)<U(f)+\frac{\epsilon}{2}\] and \[U(g;P)<U(g)+\frac{\epsilon}{2}\].

Combining all the inequalities, \[U(f+g)\le U(f+g;P)\le U(f;P)+U(g;P)<U(f)+U(g)+\epsilon\]. We get \[U(f;P)+U(g;P)<U(f)+U(g)+\epsilon\], by adding \[U(f;P)<U(f)+\frac{\epsilon}{2}\] and \[U(g;P)<U(g)+\frac{\epsilon}{2}\] together.

Since this inequality holds for any \[\epsilon>0\], then at the very least \[U(f+g)\le U(f;P)+U(g;P)\le U(f)+U(g)\]. Similarly, \[L(f+g;P)\ge L(f;P)+L(g;P)\] for all possible partitions \[P\], and for every \[\epsilon>0\], we get \[L(f+g)>L(f)+L(g)-\epsilon\], then \[L(f+g)\ge L(f)+L(g)\].

Note that \[U(f)=L(f)=\int_{a}^{b}f\], \[U(g)=L(g)=\int_{a}^{b}g\] and \[U(f+g)=L(f+g)=\int_{a}^{b}(f+g)\]. From all of our inequalities, \[U(f+g)\le U(f)+U(g)=\int_{a}^{b}f+\int_{a}^{b}g\], \[L(f+g)\ge L(f)+L(g)=\int_{a}^{b}f+\int_{a}^{b}g\]. Therefore, \[\int_{a}^{b}f+\int_{a}^{b}g\le\int_{a}^{b}(f+g)\le\int_{a}^{b}f+\int_{a}^{b}g\], and we can conclude that \[\int_{a}^{b}(f+g)=\int_{a}^{b}f+\int_{a}^{b}g\].

Suppose that \[f,g:\left[ a,b \right]\to\mathbb{R}\] are Riemann integrable and \[f\le g\], then \[\int_{a}^{b}f\le\int_{a}^{b}g\]. Note that \[f\le g\] implies that for all \[x\in \left[ a,b \right]\], \[f(x)\le g(x)\].

Suppose that \[f\ge0\] is Riemann integrable, i.e. \[f(x)\ge0\] for every \[x\in \left[ a,b \right]\]. Let \[P\] be a partition that consists only a single interval, i.e. \[P=\left\{ I_{1} \right\}\], then the lower Riemann sum based on this partition would be \[L(f;P)=\left( \inf_{\left[ a,b \right]}f \right)\cdot \left( b-a \right)\ge0\]. So, \[\int_{a}^{b}f\ge L(f;P)\ge0\].

Let function \[h=f-g\], then \[h\le0\]. Then, the linearity of integral implies that \[\int_{a}^{b}f-\int_{a}^{b}g=\int_{a}^{b}(f-g)=\int_{a}^{b}h\], and \[\int_{a}^{b}h\le0\], since \[h\le0\]. So, \[\int_{a}^{b}f-\int_{a}^{b}g\le0\], therefore, \[\int_{a}^{b}f\le\int_{a}^{b}g\], when \[f\le g\].

As a result of this inequality, we can define \[m=\sup_{\left[ a,b \right]}f\] and \[M=\inf_{\left[ a,b \right]}f\]. \[m\le f\le M\], then \[\int_{a}^{b}m\le\int_{a}^{b}f\le\int_{a}^{b}M\].

Suppose \[f:\left[ a,b \right]\to\mathbb{R}\] and \[a<c<b\]. Then \[f\] is Riemann integrable on \[\left[ a,b \right]\] if and only if its Riemann integrable on \[\left[ a,c \right]\] and \[\left[ c,b \right]\]. In that case, \[\int_{a}^{b}f=\int_{a}^{c}f+\int_{c}^{b}f\].

Suppose that \[f\] is Riemann integrable on \[\left[ a,b \right]\]. Then, for any \[\epsilon>0\], there exists a partition \[P\] of \[\left[ a,b \right]\] such that \[U(f;P)-L(f;P)<\epsilon\]. Let partition \[P^{\prime}\] be a partition obtained by combining the endpoints of \[P\] and point \[c\], or mathematically, \[P^{\prime}=P\cup \left\{ c \right\}\]. If the point \[c\] is already in \[P\], then \[P^{\prime}=P\].

Let partition \[Q\] be the intersection between \[P^{\prime}\] and \[\left[ a,c \right]\], \[Q=P^{\prime}\cap \left[ a,c \right]\], i.e. \[P^{\prime}\] contains points in \[\left[ a,b \right]\], but we only pick the points within \[\left[ a,c \right]\]. Similarly, let partition \[R=P^{\prime}\cap \left[ c,b \right]\].

Moreover, \[U(f;P^{\prime})=U(f;Q)+U(f;R)\] and \[L(f;P^{\prime})=L(f;Q)+L(f;R)\]. Then,

\begin{align*} U(f;P^{\prime})-L(f;P^{\prime})&=(U(f;Q)+U(f;R))-(L(f;Q)+L(f;R))\\ &=(U(f;Q)-L(f;Q))+(U(f;R)-L(f;R))\\ \text{Since }\,U(f;R)-L(f;R)&\ge0\\ U(f;P^{\prime})-L(f;P^{\prime})&\ge U(f;Q)-L(f;Q)\\ \end{align*}

Since partition \[P^{\prime}\] is a refinement of \[P\] (means it contains all points in \[P\] and maybe some additional points), it follows that \[U(f;P)-L(f;P)\ge U(f;P^{\prime})-L(f;P^{\prime})\].

Combining both inequalities, we get \[U(f;P)-L(f;P)\ge U(f;P^{\prime})-L(f;P^{\prime})\ge U(f;Q)-L(f;Q)\]. Then, by definition of Riemann integrable, for every \[\epsilon>0\], \[U(f;Q)-L(f;Q)\le U(f;P)-L(f;P)<\epsilon\], which proves that \[f\] is Riemann integrable on \[\left[ a,c \right]\]. We can also reach a similar conclusion that \[f\] is Riemann integrable on \[\left[ c,b \right]\].

Conversely, if \[f\] is Riemann integrable on \[\left[ a,c \right]\] and \[\left[ c,b \right]\], then there exists partitions \[Q\] of \[\left[ a,c \right]\] and \[R\] of \[\left[ c,b \right]\] such that \[U(f;Q)-L(f;Q)<\frac{\epsilon}{2}\] and \[U(f;R)-L(f;R)<\frac{\epsilon}{2}\].

Let \[P=Q\cup R\]. Then \[U(f;P)-L(f;P)=U(f;Q)-L(f;Q)+U(f;R)-L(f;R)<\frac{\epsilon}{2}+\frac{\epsilon}{2}\], \[U(f;P)-L(f;P)<\epsilon\], proving that \[f\] is Riemann integrable on \[\left[ a,b \right]\].

Finally, with all the partitions above, we have \[\int_{a}^{b}f\le U(f;P)=U(f;Q)+U(f;R)\]. Since \[U(f;Q)<L(f;Q)+\frac{\epsilon}{2}\] and \[U(f;R)<L(f;R)+\frac{\epsilon}{2}\], then \[\int_{a}^{b}f<L(f;Q)+L(f;R)+\epsilon\]. Substituting the fact that \[L(f;Q)+L(f;R)\le\int_{a}^{c}f+\int_{c}^{b}f\] into our original equation, \[\int_{a}^{b}f<\int_{a}^{c}f+\int_{c}^{b}f+\epsilon\].

Similarly, \[\int_{a}^{b}f\ge L(f;P)=L(f;Q)+L(f;R)\], \[L(f;Q)>U(f;Q)-\frac{\epsilon}{2}\] and \[L(f;R)>U(f;R)-\frac{\epsilon}{2}\], combining all of it together we get \[\int_{a}^{b}f>U(f;Q)-U(f;R)-\epsilon\]. Then, \[U(f;Q)+U(f;R)\ge\int_{a}^{c}f+\int_{c}^{b}f\], \[\int_{a}^{b}f>\int_{a}^{c}f+\int_{c}^{b}f-\epsilon\].

Now, \[\int_{a}^{b}f\] is squeezed together: \[\int_{a}^{c}f+\int_{c}^{b}f-\epsilon<\int_{a}^{b}f<\int_{a}^{c}f+\int_{c}^{b}f+\epsilon\]. Since \[\epsilon\] can be any number larger than 0, it follows that \[\int_{a}^{b}f=\int_{a}^{c}f+\int_{c}^{b}f\] (as \[\epsilon\] can get infinitely close to zero).

If \[f:\left[ a,b \right]\to\mathbb{R}\] is Riemann integrable, where \[a<b\] and \[a\le c\le b\], then \[\int_{a}^{b}f=-\int_{b}^{a}f\] and \[\int_{c}^{c}f=0\].

With this definition, we make sense of negative intervals: reversing the limits just flips the sign of the integral and integrating over an interval of zero length gives zero. This perspective ensures the standard additivity property \[\int_{a}^{b}f=\int_{a}^{c}f+\int_{c}^{b}f\] holds for any \[a,b,c\in\mathbb{R}\], without ever needing \[a<b\].