---- > [!definition] [[Riemann integral]]) > Let $Q$ be a rectangle in $\mathbb{R}^{n}$, $f:Q \to \mathbb{R}$ a [[bounded function]], let $P$ be a [[partition]] of $Q$. Intuition from calculus says that the upper sum [[upper sum]] $U(f, P)=\sum_{R \in P} \big( \sup_{x \in R} f(x) \big) v(Q)$ overestimates and the lower sum $L(f,P)=\sum_{R \in P} \big( \inf_{x \in R} f(x) \big) v(Q)$ underestimates. To this end, we defined the [[upper integral]] $\inf_{P}U(f,P)$ and [[lower integral]] $\sup_{P}L(f, P)$ by taking [[infimum|infimums]] and [[supremum|supremums]] of upper/lower sums over partitions of $Q$. > We call the bounded function $f:(Q \subset \mathbb{R}^{n}) \to \mathbb{R}$ **Riemann integrable** if these values are equal, and denote their shared value $\int _{Q} f$ or $\int _{x \in Q} f(x)$: $\overbrace{ \inf_{P}U(f,P)=\sup_{P}L(f,P) }^{ =: \int _{Q} f }.$ The number $\int _{Q} f$ is called the **Riemann integral** of $f$. > [!equivalence] > [[characterization of riemann integrability|Riemann integrable iff discontinuity set has measure zero]] ^equivalence > [!basicnonexample] Limitations of the Riemann integral. > > > The Riemann integral has several deficiencies. For instance: > - Riemann integration does not handle functions with many discontinuities, e.g. the [[Dirichlet function]] is not Riemann integrable.[^1] > - Riemann integration does not handle unbounded functions by default (but do see [[extended integral|extended Riemann integral]]) > - Riemann integration does not work well with limits [^1]: Recall that the Dirichlet function $1_{\mathbb{Q}}:[0,1] \to \mathbb{R}$ is $1_{\mathbb{Q}}(x)=\begin{cases}1 & x \text{ is rational}\\0 & x \text{ is irrational}.\end{cases}$ Since e.g. [[the rationals are dense in the reals|Q and R-Q are dense in R]], any subinterval $[a,b] \subset [0,1]$ contains a rational number and an irrational number. Hence $\inf_{x \in [a,b]}1_{\mathbb{Q}}(x)=0$ while $\sup_{x \in [a,b]}1_{\mathbb{Q}}(x)=1$. It follows that every partition $P$ of $[0,1]$, $L(1_{\mathbb{Q}}, P)=0$ while $U(1_{\mathbb{Q}}, P)=1$. Hence $L(1_{\mathbb{Q}}) \neq U(1_{\mathbb{Q}})$, so $1_{\mathbb{Q}}$ is not Riemann integrable. This is disturbing, because there are far fewer rational numbers than irrational numbers — thus $1_{\mathbb{Q}}$ should, in some sense, have integral $0$. But its Riemann integral is undefined. ---- #### ---- #### References > [!backlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM [[]] > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` > [!frontlink] > ```dataview > TABLE rows.file.link as "Further Reading" > FROM outgoing([[]]) > FLATTEN file.tags as Tag > WHERE Tag = "#definition" OR Tag = "#theorem" OR Tag = "#MOC" OR Tag = "#proposition" OR Tag = "#axiom" > GROUP BY Tag > ``` #analysis/real-analysis # Definition Let $Q$ be a [[rectangle]] in $\mathbb{R}^n$, $f: Q \to \mathbb{R}$ a [[bounded function]]. We sa[[Riemann integral|Riemann Integrable]]egrable]] if $\underline{\int_{Q}}f = \overline{\int^{Q}}f,$ (i.e., [[lower integral]]=[[upper integral]]) and write this shared value as $\int_Q f.$ Sometimes to indicate that we're only integrating over a subset of variables (e.g., when dealing with [[Fubini's Theorem]]) we use the notation $\int_{x \in Q}f(x).$ #notFormatted