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UPSC 2026 Maths Optional Paper 2 Q3b — Step-by-Step Solution

20 marks · Section A

Riemann integral · Real Analysis · asked 11× in 14 yrs · Read the full method →

Question

A function $f$ is defined on $[0, 1]$ by

Evaluate lower integral $\displaystyle\underline{\int_{0}^{1}} f$ and upper integral $\displaystyle\overline{\int_{0}^{1}} f$. Does the integral $\displaystyle\int_{0}^{1} f$ exist?

Technique

Darboux from first principles. On any subinterval both $\mathbb{Q}$ and $\mathbb{Q}^c$ are dense, so the infimum of $f$ there is the smaller of the two branch-infima and the supremum is the larger of the two branch-suprema. On $[0,1]$ we have $x^3\le x^2$ — the reverse of the ordering on $[1,\infty)$, and getting it backwards interchanges the two answers. The consequence is decisive: every lower Darboux sum of $f$ equals the corresponding lower sum of $x^3$, and every upper sum of $f$ equals the corresponding upper sum of $x^2$; so the two Darboux integrals are just $\int_0^1x^3$ and $\int_0^1x^2$. The existence question is then settled by Riemann’s criterion.

Solution

Step 1 — The pointwise ordering on $[0,1]$ (the step that decides the answer).

For $x\in[0,1]$,

with equality exactly at $x=0$ and $x=1$. Hence

since $f(x)$ is one of $x^2,x^3$ at each point. Both bounding functions are continuous and increasing on $[0,1]$ — a fact used repeatedly below.

Sanity anchor. On $[1,\infty)$ the inequality reverses ($x^3\ge x^2$). A candidate who imports that ordering will report $\underline{\int}f=\tfrac13>\overline{\int}f=\tfrac14$, which is impossible, since always $\underline{\int}f\le\overline{\int}f$. If your two numbers come out in the wrong order, the ordering of $x^2,x^3$ is where the error is.

Step 2 — Notation, and the infimum/supremum on a subinterval.

Let $P=\{0=x_0<x_1<\cdots<x_k=1\}$ be a partition of $[0,1]$, write $I_i=[x_{i-1},x_i]$, $\Delta_i=x_i-x_{i-1}>0$, and

Since $\Delta_i>0$, both $\mathbb{Q}\cap I_i$ and $\mathbb{Q}^{c}\cap I_i$ are non-empty and dense in $I_i$ (density of the rationals and of the irrationals in $\mathbb{R}$). Therefore the value set of $f$ on $I_i$ is

and consequently

The four branch extrema. Let $g$ be continuous and increasing on $I_i$ and let $D\subseteq I_i$ be dense in $I_i$. For every $x\in D$, $g(x)\ge g(x_{i-1})$, so $\inf_D g\ge g(x_{i-1})$; and by density there are $t_j\in D$ with $t_j\downarrow x_{i-1}$, so $\inf_D g\le\lim_j g(t_j)=g(x_{i-1})$ by continuity. Hence $\inf_D g=g(x_{i-1})$, and symmetrically $\sup_D g=g(x_i)$. (Neither need be attained: if $x_{i-1}$ is rational, the irrational infimum $x_{i-1}^{3}$ is a limit only.)

Applying this with $g(x)=x^{2}$ on $D=\mathbb{Q}\cap I_i$ and with $g(x)=x^{3}$ on $D=\mathbb{Q}^{c}\cap I_i$:

Now feed these into (2) and use Step 1 ($x_{i-1}^{3}\le x_{i-1}^{2}$ and $x_i^{3}\le x_i^{2}$, both endpoints lying in $[0,1]$):

Step 3 — Identify the Darboux sums of $f$ with those of $x^3$ and $x^2$.

Because $x\mapsto x^{3}$ is increasing on $[0,1]$, its infimum on $I_i$ is $x_{i-1}^{3}$; because $x\mapsto x^{2}$ is increasing, its supremum on $I_i$ is $x_i^{2}$. So (3) says precisely

Both $x^{3}$ and $x^{2}$ are continuous on $[0,1]$, hence Riemann integrable, hence their lower and upper integrals equal their ordinary integrals. Taking the supremum over $P$ in the first identity of (4) and the infimum over $P$ in the second:

Step 4 — The same two numbers computed explicitly, from a uniform partition.

(The examiner wants to see the limit done, not only quoted. This step is self-contained and re-proves Step 3’s conclusion.)

Take $P_n=\left\{0,\tfrac1n,\tfrac2n,\dots,\tfrac nn\right\}$, so $x_i=\tfrac in$ and $\Delta_i=\tfrac1n$. By (3),

Hence

By definition $\underline{\int_0^1}f\ge L(P_n,f)$ and $\overline{\int_0^1}f\le U(P_n,f)$ for each $n$, so letting $n\to\infty$,

For the reverse inequalities — the half a hurried candidate omits — use (4): for every partition $P$,

whence $\underline{\int_0^1}f\le\frac14$ and $\overline{\int_0^1}f\ge\frac13$. With (6),

(The two computations agree, and $\tfrac14<\tfrac13$ respects the general inequality $\underline{\int}\le\overline{\int}$.)

Step 5 — Does $\int_0^1 f$ exist?

Riemann’s (Darboux’s) criterion. A bounded $f$ on $[a,b]$ is Riemann integrable iff $\underline{\int_a^b}f=\overline{\int_a^b}f$; equivalently, iff for each $\varepsilon>0$ there is a partition $P$ with $U(P,f)-L(P,f)<\varepsilon$.

Here $f$ is bounded ($0\le f\le1$ on $[0,1]$ by (1)), so the criterion applies, and

Indeed, for every partition $P$ whatsoever,

so no partition can force the oscillation sum below $\varepsilon=\tfrac1{12}$.

Step 6 — Cross-check by Lebesgue’s criterion (independent second proof).

Lebesgue’s criterion. A bounded $f$ on $[a,b]$ is Riemann integrable iff its set of discontinuities has Lebesgue measure zero.

Here $f$ is continuous at $x_0\in[0,1]$ iff $x_0^{2}=x_0^{3}$, i.e. iff $x_0\in\{0,1\}$. Indeed, if $x_0^2\neq x_0^3$ then every neighbourhood of $x_0$ contains both rationals and irrationals, along which $f$ takes values near $x_0^2$ and near $x_0^3$ respectively, so $f$ has no limit at $x_0$; whereas if $x_0^2=x_0^3$ then $|f(x)-f(x_0)|\le\max(|x^2-x_0^2|,|x^3-x_0^3|)\to0$.

So the discontinuity set is $(0,1)$, of measure $1>0$, and $f$ is not Riemann integrable — agreeing with Step 5. (In the Lebesgue theory $f=x^{3}$ except on $\mathbb{Q}\cap[0,1]$, a null set, so $f$ is Lebesgue integrable with $\int_{[0,1]}f\,dm=\tfrac14$. The question, being posed in Darboux language, asks about the Riemann integral, which does not exist.)

Answer