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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.

Let $f_n$ be continuously differentiable functions on an interval $I$. If $f_n^{\prime }\to g$ by uniform convergence on $I$ and $f_n(c)\to A$ at one fixed $c\in I$, then $f_n$ converge locally uniformly to a continuously differentiable function $f$ with $f^{\prime }=g$. If $I$ is bounded, the convergence is uniform on all of $I$.^[ref]

The anchor fixes the additive constant that a derivative cannot see. By the fundamental theorem of calculus, defineThe uniform limit theorem makes $g$ a continuous function, so the fundamental theorem of calculus gives $f^{\prime }=g$. The displayed estimate proves uniform convergence wherever $|x-c|$ has a common finite bound, and therefore proves locally uniform convergence on any interval.

For example, on $[0,2]$ take $f_n(x)=3+1/n+x^2/2+x/n$. Its derivative is $x+1/n$, with supremum error $1/n$ from $g(x)=x$, and $f_n(0)\to 3$. The reconstructed uniform limit is $f(x)=3+x^2/2$; the exact supremum error is $3/n$, equal to the bound above. Without an anchor, the constant functions $f_n(x)=n$ have zero derivatives but no finite pointwise limit. On the real line, $f_n(x)=x/n$ has convergent anchored values and uniformly convergent derivatives, but its supremum error from zero is infinite. The bounded-distance condition has a real role.

A uniform limit of continuous functions is a continuous function. No compactness assumption on the domain of a function is required.^[ref]

For a proof, fix an input $x$ and tolerance $\epsilon >0$. Uniform convergence supplies one $f_N$ with $\Vert f_N-f{\Vert }_{\infty }<\epsilon /3$. Its continuity at $x$ supplies $\delta >0$ with $|f_N(y)-f_N(x)|<\epsilon /3$ for allowed inputs satisfying $|y-x|<\delta$. The triangle inequality givesOne approximating function controls both endpoints of this estimate. Under pointwise convergence, selecting an index for $x$ supplies no common control at nearby $y$. This explains why the polynomial functions $x^n$ can have a discontinuous pointwise limit on $[0,1]$.