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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 $f$.

At zero, $|f(x)-f(0)|=|x|\to 0$, so the function is continuous. At a nonzero $x_0$, approach along rational and irrational sequences; density of the rational numbers and density of the irrational numbers give limiting values $x_0$ and $-x_0$, which are different. Thus the continuity set is $\boxed{\{0\}}$.

At zero, the difference quotient is one for rational nonzero increments and minus one for irrational increments, so the derivative does not exist. Elsewhere differentiability implies continuity, and continuity already fails. Thus the differentiability set is $\boxed{\varnothing }$.

For the first additional construction, modify the rational/irrational example toAt zero, $|g(h)/h|=|h|\to 0$, so $g^{\prime }(0)=0$. At any nonzero point, density of the rational numbers and density of the irrational numbers give incompatible limiting values $x^2$ and $-x^2$. Thus $g$ is discontinuous, and hence not differentiable, there. Its differentiability set is exactly $\{0\}$.

For the second construction, use a summable absolute-value cusp series:On $[-R,R]$, each summand is bounded by $2^{-n}(R+1/2)$. The Weierstrass M-test therefore gives uniform convergence on every bounded interval, and the uniform limit theorem makes $G$ continuous everywhere.

To justify differentiability, each summand's difference quotient has absolute value at most $2^{-n}$, by the reverse triangle inequality. The sum of these bounds is finite, uniformly in the increment. Thus the tail of the difference quotient is uniformly small, and limits may be passed through the sum by first retaining finitely many terms and then making the tail small. If $x\ne 1/n$ for every $n\ge 2$, each individual summand is differentiable, givingIn particular $G^{\prime }(0)=-{\sum }_{n=2}^{\infty }{2}^{-n}=-1/2$. The accumulation of cusp locations at zero does not destroy the derivative, because their weights are summable.

At $x=1/m$, the same tail argument applies to both one-sided derivatives. Every term except the $m$th has the same derivative from both sides; the $m$th has derivatives $-2^{-m}$ and $2^{-m}$. ConsequentlyTherefore $G$ fails to be differentiable exactly at $1/2,1/3,1/4,\dots$, and is differentiable everywhere else.