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If $f,g:[a,b]\to \mathbb{R}$ are continuous functions, $g\ge 0$, and $f$ is a nondecreasing monotonic function, there is $d\in [a,b]$ satisfying the identity in the title. It expresses the weighted integral as a mixture of the two endpoint values of $f$, with the split occurring in the accumulated weight.

Put $G={\int }_a^{b}g$ and $I={\int }_a^{b}fg$. The monotonicity of the Riemann integral gives $f(a)G\le I\le f(b)G$. The function $H(x)={\int }_a^{x}g$ is continuous; indeed, $|H(y)-H(x)|\le {\max }_{[a,b]}|g||y-x|$ by the extreme value theorem and the triangle inequality for the Riemann integral. Thereforeis continuous, with $K(a)=f(b)G$ and $K(b)=f(a)G$. The intermediate value theorem gives $K(d)=I$. If the endpoint values coincide, the bounds force $I$ to have that common value and an endpoint works. The proof covers zero total weight and a degenerate interval without division.

For a continuous convex function on $[a,b]$, where $a<b$, its integral average lies between its midpoint value and the average of its endpoint values. For a concave function both inequalities reverse. The bound involving endpoints follows by comparison with the affine function joining them and monotonicity of the Riemann integral. The midpoint bound follows from the Jensen inequality on equal-weight midpoint tags and the Riemann sum convergence theorem.

Suppose the continuous nonnegative function has $f(t_0)>0$ somewhere. Continuity supplies a positive-length interval within $[a,b]$ on which $f\ge f(t_0)/2$; at an endpoint use a one-sided interval. By monotonicity of the Riemann integral, its contribution to the integral is strictly positive, a contradiction. Therefore $\boxed{f\equiv 0}$ under the stated continuous-function assumptions.

For the general integrable case, choose $c\in (a,b)$ and let $f(c)=1$ while $f(t)=0$ for $t\ne c$. This nonnegative indicator function is Riemann integrable with integral zero: isolate $c$ in partition intervals of arbitrarily small total length, making the upper sum arbitrarily small while the lower sum stays zero. Yet $f(c)=1$. This is the single-point changes preserve Riemann integrals phenomenon and supplies the counterexample.

For a partition $a=x_0<\cdots <x_k=b$ and tags ${\xi }_j\in [x_{j-1},x_j]$, form the Riemann sum ${\sum }_{j=1}^{k}f({\xi }_j)(x_j-x_{j-1})$. The Riemann integral is the unique real number $I$ such that for every $\epsilon >0$ there is $\delta >0$ for which every tagged partition with mesh ${\max }_j(x_j-x_{j-1})<\delta$ has its sum within $\epsilon$ of $I$. Then $I={\int }_a^{b}f(x)dx$. The stated continuity ensures existence, which need not be proved here.

The properties required are linearity, the integral of a constant, and monotonicity of the Riemann integral. Monotonicity follows directly because a nonnegative function has nonnegative Riemann sums, and therefore a nonnegative integral. Integrating $f-m\ge 0$ and $M-f\ge 0$ givesSince $g\ge 0$, multiplication preserves the pointwise inequalities: $mg\le fg\le Mg$. The continuous products are integrable, so the same monotonicity and linearity giveNeither $m$ nor $f$ needs to be nonnegative for this argument.

For the first limit, set $r_n=n^{-1/2}$, $g_n(x)=n{e}^{-nx}$, and ${\epsilon }_n={\max }_{0\le x\le r_n}|f(x)-f(0)|$. By continuity at zero, ${\epsilon }_n\to 0$. On this shrinking interval $f(0)-{\epsilon }_n\le f(x)\le f(0)+{\epsilon }_n$, andApplying the weighted bounds just proved givesBoth bounding expressions tend to $f(0)$, so the squeeze theorem proves the first limit.

For the full interval, continuity on $[0,1]$ gives a finite bound $B={\max }_{[0,1]}|f|$ by the extreme value theorem. The remaining tail has absolute value at mostAdding this to the already evaluated near-end integral provesThese are instances of a one-sided exponential approximate identity: the positive weight has total mass tending to one, concentrated in a shrinking neighborhood of zero. No interchange of limit and integral has been assumed.

Pointwise on $[a,b]$,because $g(x)>0$. To see the implication directly from the definition, every Riemann sum of the nonnegative continuous function $(f-m)g$ is nonnegative, so its limit is nonnegative:Applying the same argument to $(M-f)g$ gives ${\int }_a^b(M-f)g\ge 0$. By linearity, these are exactly the monotonicity of the Riemann integral bounds

Write $G={\int }_a^{b}g(t)dt\ge 0$ and $I={\int }_a^{b}f(t)g(t)dt$. By the extreme value theorem, $f$ attains a minimum $m$ and a maximum $M$ on $[a,b]$. Because $g\ge 0$, the monotonicity of the Riemann integral givesIf $G=0$, this forces $I=0$, and every $c\in [a,b]$ satisfies the required identity. This includes $a=b$.

If $G>0$, divide to obtain $m\le I/G\le M$. The intermediate value theorem, applied between points where the continuous function $f$ attains $m$ and $M$, supplies $c\in [a,b]$ with $f(c)=I/G$. Hence the weighted mean value theorem for integrals gives

Again let $G={\int }_a^{b}g(t)dt\ge 0$ and $I={\int }_a^{b}f(t)g(t)dt$. Since $f$ is a nondecreasing monotonic function, $f(a)\le f(t)\le f(b)$ on $[a,b]$. The condition $g\ge 0$ and the monotonicity of the Riemann integral giveSet $H(x)={\int }_a^{x}g(t)dt$. By part (a), $H$ is continuous. Therefore the functionis continuous on $[a,b]$. Its endpoint values are $K(a)=f(b)G$ and $K(b)=f(a)G$, which enclose $I$. The intermediate value theorem provides $d\in [a,b]$ with $K(d)=I$, soIf the two endpoint values coincide, the preceding bounds force $I$ to equal that common value and either endpoint works. Thus the proof also covers $G=0$, $f(a)=f(b)$ and $a=b$, without dividing by any possibly zero quantity.

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.

The stated weighted inequality is the Jensen inequality for a concave function. Assume $a<b$; if $a=b$, both sides of the required inequality are zero. Write $x=\lambda a+(1-\lambda )b$ with $\lambda =(b-x)/(b-a)$. Applying the weighted inequality at the two endpoints givesThus the affine function $\ell$ joining the endpoint values lies below $f$. By monotonicity of the Riemann integral,This is the lower bound in the Hermite-Hadamard inequality for a concave function.

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.