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Let $f$ be bounded on $[a,b]$ and continuous on $(a,b]$. For any positive error tolerance, isolate $[a,a+\delta ]$. Its contribution to the difference of upper Darboux sums and lower Darboux sums is at most $2\Vert f{\Vert }_{\infty }\delta$. On $[a+\delta ,b]$, uniform continuity supplies a partition of an interval with arbitrarily small remaining oscillation sum. The Riemann integrability criterion then proves integrability, even when the values oscillate without a limit at the endpoint.

For a bounded real function, inserting partition points increases its lower Darboux sum and decreases its upper Darboux sum. This follows interval by interval because infima increase on smaller sets while suprema decrease, and the new widths sum to the original width. A common partition refinement consequently proves $s(f,D_1)\le S(f,D_2)$ for arbitrary two partitions.

Let $K$ consist of numbers whose decimal digits all lie in a fixed set $D\subseteq \{1,\dots ,8\}$ with $m<10$ digits. The $m^N$ allowed prefixes give closed decimal intervals of length $10^{-N}$ covering $K$, with no endpoints in $K$. A partition of an interval at those endpoints gives an upper Darboux sum at most $(m/10{)}^N$. Every cell contains a terminating decimal outside $K$, so the lower Darboux sum is zero. Thus the indicator function is Riemann integrable with Riemann integral zero. Such sets illustrate how an uncountable set may have an integrable indicator function of zero integral.

For a nonnegative measurable function $f:X\to [0,\infty ]$ on a measure space, defineThe supremum may be infinite. A simple function assigns finitely many heights to measurable sets, so its integral is height times measure, summed over those pieces. This definition groups points by their values rather than requiring a partition into short intervals. See MIT lecture 11, Definition 108.

A real-valued function is Lebesgue integrable when ${\int }_X|f|d\mu <\infty$. Its integral is ${\int }_X{f}_{+}d\mu -{\int }_X{f}_{-}d\mu$, where $f_{+}=\max (f,0)$ is its positive part and $f_{-}=\max (-f,0)$. This finiteness requirement avoids subtracting two infinities. For functions taking values in the complex numbers, integrate the real part and imaginary part separately.

Consider the Dirichlet function $f=1_{\mathbb{Q}\cap [0,1]}$ on $[0,1]$. It is a simple function, andIts Riemann integral does not exist. For every partition of an interval $0=x_0<\cdots <x_n=1$, each nondegenerate $[x_{i-1},x_i]$ contains both rational numbers and irrational numbers, by their respective density of the rational numbers and density of the irrational numbers. Thus its infimum there is $0$ and its supremum is $1$. The lower Darboux sum is ${\sum }_{i}0(x_i-x_{i-1})=0$, whereas the upper Darboux sum is ${\sum }_{i}1(x_i-x_{i-1})=1$. Refinement never closes this gap. A finite scatterplot cannot represent this behavior: the assertion concerns every subinterval and every partition of an interval.

Another view builds the Lebesgue integral from horizontal layers. For any nonnegative measurable function, the layer cake representation statesBoth sides may be infinite. The right side integrates the measure of a superlevel set against Lebesgue measure in the height variable $t$. See Gautam Iyer's measure theory notes, Theorem 9.5.

For $f(x)=x^2$ on $[0,1]$ with Lebesgue measure, the superlevel set at $0\le t<1$ is $(\sqrt{t},1]$, of Lebesgue measure $1-\sqrt{t}$; for $t\ge 1$ it is empty. ConsequentlyAt height $t=1/4$, the layer extends from $x=1/2$ to $x=1$, a width of $1/2$. The picture shows why the same area can be accumulated vertically or horizontally. In a probability space, the identical formula expresses the expected value of a nonnegative random variable through its exceedance probabilities.

A proof outline avoids extra assumptions on the measure space. For a nonnegative simple function $s={\sum }_j{a}_j{1}_{A_j}$ with disjoint measurable sets $A_j$, the superlevel set has measure ${\sum }_{a_j>t}\mu (A_j)$. Integrating this finite sum gives ${\sum }_j{a}_j\mu (A_j)=\int sd\mu$. Next use approximation by nonnegative simple functions $s_n\uparrow f$. For every $t$, the superlevel sets $\{s_n>t\}$ increase to $\{f>t\}$. Continuity from below of a measure and the monotone convergence theorem on the height variable give the formula for $f$.

The lower Darboux integral of a bounded real function on a closed interval is the supremum of its lower Darboux sums over all partitions of an interval. Refinement increases a lower Darboux sum. A common refinement of two arbitrary partitions of an interval proves that every lower Darboux sum is at most every upper Darboux sum, hence the lower Darboux integral never exceeds the upper Darboux integral.

For a bounded real function, its oscillation on an interval is its full range width there. An upper Darboux sum minus a lower Darboux sum is the sum of these widths multiplied by their interval lengths. This gives the Riemann integrability criterion in terms of oscillation.

For a bounded function on an interval $I$, write ${\omega }_I(f)={\sup }_{I}f-{\inf }_{I}f$ for its oscillation of a function on an interval. For a partition $P$, the difference between its upper Darboux sum and lower Darboux sum is ${\sum }_{I\in P}|I|{\omega }_I(f)$. The Riemann integrability criterion says that this can be made arbitrarily small.

Since ${\omega }_I(f_1+f_2)\le {\omega }_I(f_1)+{\omega }_I(f_2)$, choose good partitions for the two functions and take their common refinement. Darboux sum refinement monotonicity makes both oscillation sums small on this refinement, so their sum is Riemann integrable. The same argument handles differences and scalar multiples.

Both $t\mapsto \max (t,0)$ and $t\mapsto |t|$ are Lipschitz continuous with constant one. Thus ${\omega }_I(f^{+})\le {\omega }_I(f)$ and ${\omega }_I(|f|)\le {\omega }_I(f)$. This calculation shows that Lipschitz composition preserves Riemann integrability, so the positive part and the absolute value are integrable.

The converse for the absolute value is false. Define $f=1$ on rational numbers and $f=-1$ on irrational numbers. Then $|f|=1$ is integrable, but every interval has supremum one and infimum minus one, so every Darboux difference for $f$ is $2(b-a)$. Hence $f$ is not Riemann integrable.

Finally,is integrable by the closure properties just proved. This establishes that Riemann integrability is closed under addition and maximum.

A countable supremum can fail, even for a uniformly bounded family. Enumerate the rational points of $[a,b]$ as $q_1,q_2,\dots$ and set $f_n=1_{\{q_n\}}$. Each singleton indicator function is Riemann integrable with integral zero: isolate its single nonzero point in intervals of arbitrarily small total length. Butis the nonintegrable Dirichlet function. This proves that a pointwise supremum need not preserve Riemann integrability.

The PDF defines this function using $\sin (1/x)$; the converted TeX contains a conflicting duplicate $\sin (1/2)$ before its later correct version. Use the PDF function. It is bounded by one and continuous at every positive $x$, although it is discontinuous at zero. Indeed the sequences $x_n=(\pi /2+2\pi n{)}^{-1}$ and $y_n=(3\pi /2+2\pi n{)}^{-1}$ tend to zero with function values $1$ and $-1$.

That single discontinuity does not prevent Riemann integrability. Given $\epsilon >0$, choose $0<\delta <\min \{1,\epsilon /4\}$. The cell $[0,\delta ]$ contributes at most $2\delta <\epsilon /2$ to the difference between the upper Darboux sum and the lower Darboux sum. On $[\delta ,1]$ the function is uniformly continuous, so choose a partition there whose oscillation sum is less than $\epsilon /2$, by the introductory proof. Joining that partition to the cell $[0,\delta ]$ givesThus the PDF's function is Riemann integrable despite its endpoint discontinuity. This is a direct application of bounded functions continuous away from one endpoint are Riemann integrable, proved here with the actual oscillation bound rather than an improper-integral argument.

For a bounded function, a partition of an interval is $P:0=x_0<x_1<\cdots <x_N=1$. Put $m_j={\inf }_{[x_{j-1},x_j]}f$ and $M_j={\sup }_{[x_{j-1},x_j]}f$. Its lower Darboux sum and upper Darboux sum areThe lower Darboux integral is ${\sup }_{P}L(f,P)$ and the upper Darboux integral is ${\inf }_{P}U(f,P)$. Riemann integrability means that these two real numbers coincide, and their common value is the Riemann integral.

Equivalently, the Riemann integrability criterion says that for every $\epsilon >0$ some partition satisfies $U(f,P)-L(f,P)<\epsilon$. To see equivalence, all lower sums are at most all upper sums: refine any two partitions to their common refinement and use Darboux sum refinement monotonicity. Thus a small gap forces the difference of the two integrals below $\epsilon$. Conversely, if both integrals equal $I$, choose a lower sum above $I-\epsilon /2$ and an upper sum below $I+\epsilon /2$; their common refinement has gap below $\epsilon$.

This also agrees with the tagged Riemann sum definition: there is $I$ such that every tagged sum on a partition of sufficiently small mesh is within any prescribed tolerance of $I$. Each tagged sum lies between the two Darboux sums. From a partition with small gap, an arbitrary small-mesh partition differs from its refinement at only the cells crossing the finitely many original division points; boundedness makes their total contribution tend to zero. Conversely, on a fine partition tags approximating each cell supremum or infimum force both Darboux sums close to the same tagged-sum limit.

Now suppose $f$ is continuous on $[0,1]$. It is bounded by the result of Question 11. It is also uniformly continuous: otherwise there are $x_n,y_n\in [0,1]$ with $|x_n-y_n|\to 0$ but $|f(x_n)-f(y_n)|\ge {\epsilon }_0>0$. The Bolzano-Weierstrass theorem gives a convergent subsequence of $x_n$; the corresponding $y_n$ have the same limit, contradicting continuity. Given $\epsilon >0$, this uniform continuity gives $\delta >0$ such that points less than $\delta$ apart have function values less than $\epsilon /2$ apart. A partition of mesh below $\delta$ then satisfiesHence every continuous function on the interval is Riemann integrable, with the required proof supplied by its oscillation estimates.

For a partition $D=\{a=x_0<x_1<\cdots <x_n=b\}$, put $\Delta x_k=x_k-x_{k-1}$ andBoundedness makes these finite. The lower Darboux sum and upper Darboux sum areSince $m_k\le M_k$ and the widths are positive, $s(f,D)\le S(f,D)$.

Splitting a partition interval into smaller intervals can only increase each infimum and decrease each supremum. The smaller widths sum to the original width. Its new lower contribution is therefore at least its old lower contribution, and its new upper contribution at most its old upper contribution. Repeating this for every inserted point proves Darboux sum refinement monotonicity:For arbitrary partitions, take their common partition refinement $D_{\ast }=D_1\cup D_2$. ThenNo compatibility of the original partitions is required.

Finally suppose $f\ge 0$ is Riemann integrable. Then $t_k=m_k\Delta x_k\ge 0$, so every product factor is nonnegative and the exponential inequality can be multiplied safely:A lower Darboux sum is no greater than the Riemann integral, since the integral equals the supremum of all lower sums. Monotonicity of the exponential therefore gives the exponential bound for a lower-sum productFor a degenerate interval $a=b$, the empty product and the exponential are both one.

For a bounded real function on $[0,1]$, take a partition $P$ with nodes $0=x_0<\cdots <x_N=1$. Its lower Darboux sum and upper Darboux sum areThe function is Riemann integrable when ${\sup }_{P}L(f,P)={\inf }_{P}U(f,P)$; their common value is its Riemann integral. Equivalently, for every $\eta >0$ some partition has $U-L<\eta$. Indeed a common refinement of nearly extremizing upper and lower partitions proves necessity, while arbitrarily small gaps force equality of the two integrals. This is the Riemann integrability criterion.

A continuous function on a compact interval is bounded and uniformly continuous. Choose a mesh small enough that its oscillation on each subinterval is less than $\eta$. Then $U-L\le \eta {\sum }_j(x_j-x_{j-1})=\eta$; using $\eta /2$ first makes the desired inequality strict. Thus every continuous function on $[0,1]$ is Riemann integrable.

For a partition $P$, write $U(f,P)$ and $L(f,P)$ for the upper Darboux sum and lower Darboux sum. The Riemann integrability criterion for bounded $f$ is that for every $\epsilon >0$ some partition satisfies $U(f,P)-L(f,P)<\epsilon$. Thus if the gaps for $D_n$ tend to zero, the Riemann integrability criterion immediately proves that $f$ is Riemann integrable.

Conversely, suppose $f$ is Riemann integrable, and choose a fixed partition $P$ with $U(f,P)-L(f,P)<\epsilon /2$. Let $r$ be the number of its interior division points and choose $M\ge 1$ with $|f|\le M$. Call a cell of $D_n$ bad when its interior contains a division point of $P$. There are at most $r$ bad cells, and their total length is at most $r/n$.

Every other cell lies in a single cell of $P$, so its oscillation is no larger than that of the containing cell. Summing the contributions of these good cells gives at most $U(f,P)-L(f,P)$, while each bad cell has oscillation at most $2M$. Therefore the finite bad-cell estimate for Darboux sums givesFor large enough $n$, the right-hand side is less than $\epsilon$. This proves the uniform-mesh Darboux criterionThis argument does not assume that the partitions $D_n$ are nested; in general they are not.

For the composition, $f$ takes values in $[-M,M]$. Since $g$ is continuously differentiable, the extreme value theorem bounds both $|g|$ and $|g^{\prime }|$ on that interval. In particular, let $K={\sup }_{[-M,M]}|g^{\prime }|<\infty$. The mean value theorem states that a function continuous on the interval between $s,t$ and differentiable in its interior satisfies $g(s)-g(t)=g^{\prime }(\xi )(s-t)$ for some intermediate $\xi$. HenceOn each partition cell, this bounds the oscillation of $g\circ f$ by $K$ times the oscillation of $f$, whether or not the local extrema are attained. It follows thatThe composition is bounded, so the proved uniform-mesh Darboux criterion applies. Thus $g\circ f$ is Riemann integrable. The reusable fact is Lipschitz composition preserves Riemann integrability; continuous differentiability supplies the required bound on the range of $f$.

Let $L$ be the supremum of all lower Darboux sums and $U$ the infimum of all upper Darboux sums. The assumed comparison of arbitrary lower and upper sums gives $L\le U$. For the partition supplied for a given $\epsilon$,Since this holds for every positive $\epsilon$, $U=L$. Thus the Riemann integrability criterion gives

If $f$ is continuous on $[a,b]$, the Heine-Cantor theorem makes it uniformly continuous. Given $\epsilon >0$, choose $\delta$ so that intervals of length below $\delta$ have oscillation below $\epsilon /(b-a)$. A partition of mesh below $\delta$ then satisfiesproving that Continuous functions are Riemann integrable.

For the function defined using $g$, choose $K$ with $|f|\le K$. Given $\epsilon >0$, choose $\delta >0$ so small that $2K\delta <\epsilon /2$. On $[\delta ,1]$, the function $g$ is continuous and therefore has a partition whose upper-minus-lower sum is below $\epsilon /2$. Adding the interval $[0,\delta ]$ contributes at most $2K\delta$, regardless of the value $f(0)=\lambda$. Hence the full Darboux-sum difference is below $\epsilon$, and

Finally, for any partition $a=x_0<\cdots <x_n=b$, the mean value theorem supplies ${\xi }_i\in (x_{i-1},x_i)$ such thatSumming telescopes toa Riemann sum for the Riemann-integrable function $f^{\prime }$. Equivalently, every lower derivative sum is at most this telescoping value and every upper derivative sum is at least it. Taking the common upper and lower Darboux integrals proves the fundamental theorem of calculus conclusion

For a partition of an interval $P:0=x_0<\cdots <x_m=1$, putThe upper Darboux sum and lower Darboux sum areThe upper and lower integrals areThe bounded function is Riemann integrable when these values agree, and their common value is its Riemann integral.

For the indicator of the rational numbers, every nondegenerate interval contains both a rational number and an irrational number. Hence every upper sum is one and every lower sum is zero. The function is not Riemann integrable.

Put $E_n={\sup }_{t\in [a,b]}|f(t)-{\varphi }_n(t)|\to 0$, so the hypothesis is uniform convergence. Every step function is bounded and Riemann integrable: away from its finitely many partition points it is constant, and those exceptional points can be covered by intervals of arbitrarily small total length. A partition of an interval containing their boundaries makes the difference of its upper Darboux sum and lower Darboux sum arbitrarily small.

Choose one $n$ with $E_n<\infty$. Boundedness of ${\varphi }_n$ and $|f|\le |{\varphi }_n|+E_n$ show that $f$ is a bounded function. Assume $a<b$, since for $a=b$ all the integrals are zero. Given $\eta >0$, choose $n$ so that $2E_n(b-a)<\eta /2$, and then a partition of an interval $P$ such that $U({\varphi }_n,P)-L({\varphi }_n,P)<\eta /2$. On every partition subinterval, $|f-{\varphi }_n|\le E_n$ givesThusThe Riemann integrability criterion proves that $f$ is Riemann integrable. Finally, linearity of the integral and monotonicity of the Riemann integral giveThereforeThis proves the relevant case of uniform convergence preserves integrals on a compact interval.

If Riemann integrable functions $f_n:[a,b]\to \mathbb{R}$ converge by uniform convergence to $f$, then $f$ is Riemann integrable andThis concerns one fixed compact interval, with finite length.^[ref]

Here is a proof outline that also establishes Riemann integrability of the uniform limit. Assume $a<b$, and put $\ell =b-a$ and $E_N=\Vert f_N-f{\Vert }_{\infty }$. One bounded $f_N$ and finite $E_N$ show that $f$ is a bounded function. For any partition of an interval $P$, its upper Darboux sum and lower Darboux sum satisfyGiven $\eta >0$, choose $N$ with $E_N<\eta /(4\ell )$, then a partition of an interval with $U(f_N,P)-L(f_N,P)<\eta /2$. The displayed gap is below $\eta$, proving Riemann integrability. Finally, $|f_n-f|\le E_n$ and monotonicity of the Riemann integral give the stated error estimate. When $a=b$, every integral is zero.

For the polynomial functions $f_n(x)=x^n$ on $[0,0.8]$, direct integration and the uniform bound giveAt $n=21$, the exact integral is approximately $0.0003353953$, while the general error bound is approximately $0.0073786976$. The bound certifies the result without an exact antiderivative. On $[0,1]$, these same integrals still tend to zero, although there is no uniform convergence: sufficient conditions need not be necessary. On an unbounded interval, even uniform convergence can fail to control total area, as uniformly vanishing functions can retain a nonzero integral shows.

Uniform convergence alone also fails to preserve derivatives: $f_n(x)=\sin (n^{2}x)/n$ converges uniformly to zero on $[-1,1]$, while $f_n^{\prime }(0)=n$. The small amplitude does not control the slopes of the sine oscillations.

A bounded function on $[0,1]$ is Riemann integrable if and only if its upper Darboux sum minus its lower Darboux sum on the equal-length partition $D_n$ tends to zero. Sufficiency follows directly from the Riemann integrability criterion. Necessity follows by comparing with a fixed partition whose gap is small and using the finite bad-cell estimate for Darboux sums. The equal-length partitions need not be nested.