Part 1: The Algebraic & Topological Foundations

A rigorous exploration of pointwise vs. interval monotonicity, Darboux's Theorem, Froda's Theorem, and the essential manipulation techniques that separate the elite from the rest.

In the conventional calculus track, monotonicity is reduced to a sign check: if $f'(x) > 0$ , the function increases. For the aspirant targeting the top 0.01% — JEE Advanced, ISI, CMI, or Putnam — this simplification is dangerously incomplete. Functions may be strictly increasing at a point while decreasing in every neighborhood of that point. They may be continuous and monotone yet have a derivative of zero almost everywhere.

To master monotonicity at this level is to master the tension between algebraic rigidity and topological flexibility.

I. What Does “Increasing” Actually Mean?

The most persistent misconception in competitive mathematics — and the source of many counterexamples on the Putnam — is the conflation of “increasing at a point” with “increasing on an interval.”

Definition: Monotonicity at a Point

A function $f$ is said to be increasing at a point $c$ if there exists a neighborhood $(c - \delta, c + \delta)$ such that:

If $x < c$ , then $f(x) \le f(c)$ .
If $x > c$ , then $f(x) \ge f(c)$ .

Strict monotonicity replaces $\le$ and $\ge$ with $<$ and $>$ .

Definition: Monotonicity on an Interval

A function $f$ is increasing on an interval $I$ if for every pair $x_1, x_2 \in I$ with $x_1 < x_2$ , we have $f(x_1) \le f(x_2)$ .

A fundamental result: if a function is increasing at every point in an open interval, it is increasing on the interval. However, the converse relationship via derivatives is where the traps lie.

The Positive Derivative Trap

Critical Misconception

It is true that if $f'(x) > 0$ for all $x \in (a,b)$ , then $f$ is strictly increasing on $(a,b)$ .

It is false that if $f'(c) > 0$ at a specific point $c$ , then $f$ is increasing in some neighborhood of $c$ .

Consider this pathological function, a favourite in advanced analysis courses:

f(x) = \begin{cases} x + 2x^2 \sin\left(\frac{1}{x}\right) & x \neq 0 \\ 0 & x = 0 \end{cases}

At $x = 0$ , the derivative via first principles:

f'(0) = \lim_{h \to 0} \frac{h + 2h^2 \sin(1/h)}{h} = \lim_{h \to 0} (1 + 2h \sin(1/h)) = 1 > 0

Strictly positive. Yet for $x \neq 0$ :

f'(x) = 1 + 4x\sin(1/x) - 2\cos(1/x)

As $x \to 0$ , the term $-2\cos(1/x)$ oscillates violently between $-2$ and $2$ . The derivative $f'(x)$ frequently becomes negative arbitrarily close to zero. The function is not increasing on any open interval containing $0$ , despite $f'(0) > 0$ .

Why 'f'(c) > 0 ⟹ Increasing Near c' Fails

The intuition ” $f'(c) > 0$ means the function is rising, so it must go up near $c$ ” conflates instantaneous rate with interval behavior. The derivative $f'(c) > 0$ guarantees $f(x) < f(c)$ for $x$ slightly less than $c$ and $f(x) > f(c)$ for $x$ slightly greater — but only relative to $c$ . It does not preserve the order between two arbitrary points $x_1 < x_2$ near $c$ . The oscillation of $f'$ can reverse the ordering between nearby points even while the function climbs through $c$ itself.

So when does a vanishing derivative not destroy strict monotonicity? The answer is beautiful in its simplicity:

The Discrete Zero Theorem

$f$ is strictly increasing on $[a, b]$ if $f'(x) \ge 0$ for all $x \in [a,b]$ AND the set of points where $f'(x) = 0$ is discrete (does not contain any interval).

Stationary points do not break strict monotonicity unless they form a plateau. This distinction — between an isolated zero and a flat segment — is the key shortcut.

Problem Easy

Let $f(x) = x^3$ . Is $f$ strictly increasing on $\mathbb{R}$ ?

View Solution

$f'(x) = 3x^2 \ge 0$ for all $x \in \mathbb{R}$ .
$f'(x) = 0$ only at $x = 0$ — a single point, which is a discrete set.
Since $\{0\}$ does not contain any interval, the Discrete Zero Theorem applies.
Conclusion: $f$ is strictly increasing on $\mathbb{R}$ .

This seemingly trivial result illustrates a critical principle. The same argument extends to $f(x) = x^5$ , $f(x) = x^{2k+1}$ , and many composite functions where the derivative vanishes at isolated points.

II. Darboux’s Theorem: The Hidden Power of Derivatives

A common misconception is that derivatives must be continuous. They need not be. But Jean Gaston Darboux proved that they possess something almost as powerful.

Darboux's Theorem (The Intermediate Value Property of Derivatives)

Let $f$ be a differentiable real-valued function on $[a, b]$ . If $k$ is any value between $f'(a)$ and $f'(b)$ , then there exists some $c \in (a, b)$ such that $f'(c) = k$ .

In elementary calculus, the Intermediate Value Theorem applies to continuous functions. Darboux shows that derivatives — even wildly discontinuous ones — automatically satisfy the IVT. This has profound consequences:

Exploiting Darboux in Competitions

1. Non-Jump Discontinuities: A derivative $f'$ cannot have a simple jump discontinuity. If $f'$ exists everywhere, it cannot jump from 1 to 2 without passing through all intermediate values. The only possible discontinuities of a derivative are oscillatory (like $\sin(1/x)$ ).

2. Sign Preservation: If $f'(x) \neq 0$ for all $x \in (a,b)$ , then $f'$ must maintain a constant sign throughout $(a,b)$ . This allows us to conclude strict monotonicity by checking $f'$ at a single convenient point, provided we can show $f'$ never vanishes. This technique appears routinely in ISI/CMI subjective problems.

3. Root Counting: Combined with Rolle’s Theorem, if $f'$ takes values of opposing sign, an intermediate zero must exist — even if $f'$ is discontinuous.

Trap: Derivative Implies Global Trend

Error: Assuming $f'(x) > 0$ everywhere implies $f(x) \to \infty$ .

Correction: $f(x) = \arctan(x)$ has $f'(x) > 0$ everywhere but is bounded. Monotonicity implies direction, not unbounded growth. Always check for asymptotes.

Problem Advanced

Is there a strictly increasing function $f: \mathbb{R} \to \mathbb{R}$ such that $f'(x) = f(f(x))$ for all $x$ ?

View Solution

Step 1 — Initial Deductions: If $f$ is strictly increasing, then $f' = f \circ f$ must be non-negative, so $f'(x) \ge 0$ . Since $f$ is strictly increasing, $f \circ f$ is also strictly increasing (composition of increasing functions), so $f'$ is increasing. This implies $f$ is convex.

Step 2 — Growth Analysis: If $f(x) \ge cx$ for large $x$ (with $c > 0$ ), then:

f'(x) = f(f(x)) \ge f(cx) \ge c \cdot cx = c^2 x

This implies super-linear growth in $f'$ , hence super-quadratic growth in $f$ , which feeds back into even faster growth of $f'$ . The growth rate escalates without bound.

Step 3 — Rigorous Contradiction: Assume $f(x) > x$ for large $x$ . Then $f(f(x)) > f(x) > x$ , so $f'(x) > x$ . Integrating:

f(x) > \frac{x^2}{2} + C \quad \text{for large } x

But then $f(f(x)) > f(x^2/2) > (x^2/2)^2/2 = x^4/8$ , so $f'(x) > x^4/8$ . Integrating yields $f(x) > x^5/40$ , making $f(f(x))$ grow faster still. This cascade produces a contradiction — no smooth function can sustain this feedback loop.

Step 4 — The Slow Growth Case: If $f(x) \le x$ for large $x$ , similar bounding arguments show $f'$ becomes too small to equal $f(f(x))$ .

Conclusion: No such function exists.

20 min Putnam 2010, Problem B5

III. The Domain Paradox: Disjoint Intervals

When a function’s domain is a union of disjoint intervals, the standard derivative test becomes dangerously incomplete. This is one of the most commonly missed traps in JEE Advanced.

The Jump Condition for Disjoint Intervals

When a domain is $I_1 \cup I_2$ with $I_1$ to the left of $I_2$ , the function $f$ is increasing on $I_1 \cup I_2$ if and only if:

$f$ is increasing on $I_1$ ,
$f$ is increasing on $I_2$ , and
The Jump Condition: $\sup_{I_1} f \le \inf_{I_2} f$ .

For piecewise functions or unions of intervals, explicitly checking boundary values is a mandatory step that most students skip.

The classic trap: $f(x) = 1/x$ on $(-1, 0) \cup (0, 1)$ . We have $f'(x) = -1/x^2 < 0$ everywhere — the derivative is negative throughout. But $f(-0.5) = -2$ and $f(0.5) = 2$ , so $f(-0.5) < f(0.5)$ despite $f' < 0$ . The function is not decreasing on its full domain.

Problem Medium

Let $f(x) = \ln|x| + \frac{1}{x}$ . Discuss the monotonicity of $f$ on its domain.

View Solution

Step 1 — Domain: $f$ is defined for $x \neq 0$ . Domain is $(-\infty, 0) \cup (0, \infty)$ .

Step 2 — Derivative:

f'(x) = \frac{1}{x} - \frac{1}{x^2} = \frac{x - 1}{x^2}

Step 3 — Sign Analysis:

$x^2 > 0$ always.
$f'(x) > 0$ when $x > 1$ , and $f'(x) < 0$ when $x < 1$ (and $x \neq 0$ ).

Step 4 — Monotonicity on Each Piece:

Decreasing on $(-\infty, 0)$ (since $x < 1$ throughout).
Decreasing on $(0, 1)$ .
Increasing on $(1, \infty)$ .

Step 5 — The Jump Condition Check: Is $f$ decreasing on $(-\infty, 0) \cup (0, 1)$ ?

$\lim_{x \to 0^-} f(x) = -\infty$ (the $1/x$ term dominates).
$\lim_{x \to 0^+} f(x) = +\infty$ .
The function “resets” from $-\infty$ to $+\infty$ across the discontinuity.

Conclusion: $f$ is decreasing on $(-\infty, 0)$ and decreasing on $(0, 1)$ separately. It is not monotonic on $(-\infty, 0) \cup (0, 1)$ .

8 min JEE Advanced Context

IV. Inequality Stripping & Functional Equations

For strictly monotonic functions, functional inequalities can be algebraically “stripped” — but only with extreme care about domains.

Inequality Stripping (Functional Nesting)

If $f$ is strictly increasing: $f(g(x)) \ge f(h(x)) \implies g(x) \ge h(x)$
If $f$ is strictly decreasing: $f(g(x)) \ge f(h(x)) \implies g(x) \le h(x)$ (inequality reversal)

Critical: Always verify the domain constraint first — $g(x)$ and $h(x)$ must both lie in the domain of $f$ before stripping. This is the step most students omit, and the step problem-setters exploit.

Problem Medium

Solve for $x$ : $f(3x + 1) \ge f(x^2 + x)$ given that $f$ is a strictly decreasing function defined on $[0, \infty)$ .

View Solution

Step 1 — Monotonicity Reversal: Since $f$ is strictly decreasing, strip and reverse:

f(3x+1) \ge f(x^2+x) \implies 3x + 1 \le x^2 + x

This gives $x^2 - 2x - 1 \ge 0$ , so $x \le 1 - \sqrt{2}$ or $x \ge 1 + \sqrt{2}$ .

Step 2 — Domain Constraints: Both inputs must lie in $[0, \infty)$ :

$3x + 1 \ge 0 \implies x \ge -1/3$ .
$x^2 + x \ge 0 \implies x \le -1$ or $x \ge 0$ .
Intersection: $x \ge 0$ .

Step 3 — Final Answer:

Algebraic solution: $x \le 1 - \sqrt{2}$ or $x \ge 1 + \sqrt{2}$ .
Domain constraint: $x \ge 0$ .
Answer: $\boxed{x \ge 1 + \sqrt{2}}$ .

Forgetting Domain Constraints

A common error is to answer ” $x \le 1 - \sqrt{2}$ or $x \ge 1 + \sqrt{2}$ ” without checking that inputs to $f$ lie in its domain $[0, \infty)$ . Since $1 - \sqrt{2} < 0$ , the entire left branch is eliminated. This is exactly the kind of trap JEE Advanced setters design.

6 min Faculty Notes

Parametric Isolation

For problems asking “find the set of values of parameter $a$ for which $f(x)$ is monotonic”:

Require $f'(x) \ge 0$ (or $\le 0$ ) for all $x$ in the domain.
Separate variables: rewrite as $a \ge g(x)$ or $a \le g(x)$ .
The condition holds iff $a \ge \max g(x)$ or $a \le \min g(x)$ .
This transforms a calculus problem into an algebraic optimisation problem.

Example: Find $a$ such that $f(x) = x^3 + ax^2 + 3x + 1$ is increasing on $\mathbb{R}$ . The condition $f'(x) = 3x^2 + 2ax + 3 \ge 0$ for all $x$ reduces to discriminant $\le 0$ : $4a^2 - 36 \le 0$ , giving $|a| \le 3$ .

The Involution Integral (ISI B.Math 2016)

A beautiful problem that combines monotonicity deduction with integration:

Problem Hard

Let $f: [0,1] \to [0,1]$ be differentiable with $f(f(x)) = x$ for all $x \in [0,1]$ and $f(0) = 1$ . Evaluate $\int_0^1 f(x) \, dx$ .

Hint: What kind of function satisfies f(f(x)) = x?

$f$ is an involution — it is its own inverse. What does this imply about monotonicity and the graph’s symmetry?

View Solution

Step 1 — Monotonicity Deduction: Since $f(f(x)) = x$ , the function $f$ is injective (one-to-one), hence strictly monotone.

Step 2 — Direction: $f(0) = 1$ . Apply $f$ : $f(1) = f(f(0)) = 0$ . Since $f(0) = 1 > 0 = f(1)$ , $f$ is strictly decreasing.

Step 3 — Symmetry: $f(f(x)) = x$ means the graph of $f$ is symmetric about the line $y = x$ .

Step 4 — The Elegant Invariance Argument: Since $f$ is its own inverse ( $f = f^{-1}$ ), we use the geometric identity: for any strictly monotone bijection $g: [0,a] \to [0,b]$ with $g(0) = b$ :

\int_0^a g(x) \, dx + \int_0^b g^{-1}(y) \, dy = ab

This is because the graph of $g$ and $g^{-1}$ together tile the rectangle $[0,a] \times [0,b]$ .

Here $a = b = 1$ and $f = f^{-1}$ , so:

2\int_0^1 f(x) \, dx = 1 \implies \boxed{\int_0^1 f(x) \, dx = \frac{1}{2}}

This result holds for any such involution — not just $f(x) = 1 - x$ .

12 min ISI B.Math 2016, Problem 7

V. The Topology of Monotone Functions

We now turn to the deep structural results that constrain how monotone functions can behave — or misbehave.

Froda’s Theorem: Counting Discontinuities

Froda's Theorem

The set of discontinuities of a monotone function defined on an interval is at most countable. Furthermore, all such discontinuities must be jump discontinuities (discontinuities of the first kind).

Proof Sketch: Let $f$ be monotonically increasing on $[a, b]$ . At each discontinuity $c$ , the left and right limits exist with $f(c^-) < f(c^+)$ . Associate each discontinuity with the open interval $(f(c^-), f(c^+))$ in the range. Since $f$ is monotone, these intervals are disjoint. Each contains a distinct rational number (by density of $\mathbb{Q}$ ), and the rationals are countable. $\blacksquare$

Competition Applications of Froda's Theorem

Impossibility Arguments: If a problem describes a monotonic function with an uncountable set of discontinuities, it cannot exist.
Almost Everywhere Continuity: Monotone functions are continuous except at countably many points. By Lebesgue’s theorem, they are differentiable almost everywhere.
Synthesis Problems: When constructing a monotone function with specific properties, Froda guarantees you can only prescribe countably many jumps.

Trap: The Integral Monotonicity Fallacy

Error: If $\int_a^x f(t) \, dt \ge 0$ for all $x$ , then $f(x) \ge 0$ everywhere.

Correction: This holds only for continuous $f$ . A discontinuous $f$ can be negative on a set of measure zero without affecting the integral. The correct conclusion is $f(x) \ge 0$ almost everywhere.

Dini’s Theorem: From Pointwise to Uniform

Dini's Theorem

If a monotone sequence of continuous functions $\{f_n\}$ converges pointwise to a continuous function $f$ on a compact set $[a, b]$ , then the convergence is uniform.

This allows the interchange of limits and integrals: $\lim \int f_n = \int \lim f_n$ . In competition problems with sequences of functions, establishing monotonicity in $n$ is often the “hidden step” required to justify this swap.

Trap: Limits Preserving Strict Inequalities

Error: If $\{f_n\}$ is a sequence of strictly increasing functions converging to $f$ , then $f$ is strictly increasing.

Correction: Limits only preserve weak inequalities. Example: $f_n(x) = x/n$ is strictly increasing for all $n$ , but $\lim f_n = 0$ is constant.

Heuristic: When taking limits of monotonic functions, always relax strict $(<)$ to weak $(\le)$ .

Sequences: Where Analysis Meets Algebra

Problem Medium

Let $a_n = n^{1/n}$ . Determine the maximum term and the limit.

View Solution

Step 1 — Continuise: Define $g(x) = x^{1/x} = e^{(\ln x)/x}$ and let $h(x) = \frac{\ln x}{x}$ .

Step 2 — Differentiate:

h'(x) = \frac{1 - \ln x}{x^2}

Step 3 — Critical Point: $h'(x) = 0 \implies x = e \approx 2.718$ .

Step 4 — Monotonicity:

$h'(x) > 0$ for $x < e$ → $g$ increases.
$h'(x) < 0$ for $x > e$ → $g$ decreases.

Step 5 — Integer Comparison: Maximum at $n = 2$ or $n = 3$ :

$a_2 = \sqrt{2} \approx 1.414$
$a_3 = 3^{1/3} \approx 1.442$
Check: $3^{1/3} > 2^{1/2} \iff 3^2 > 2^3 \iff 9 > 8$ ✓

Step 6 — Convergence: For $n \ge 3$ , the sequence is decreasing and bounded below by 1. By the Monotone Convergence Theorem:

\lim_{n \to \infty} n^{1/n} = e^{\lim (\ln n)/n} = e^0 = 1

8 min ISI B.Math / TOMATO 150

Problem Hard

Let $a_1 = a, b_1 = b$ be positive reals with $a_{n+1} = \frac{a_n + b_n}{2}$ and $b_{n+1} = \sqrt{a_n b_n}$ . Show both sequences converge to the same limit.

Hint: AM-GM

What does the AM-GM inequality tell you about $a_{n+1}$ vs $b_{n+1}$ ? Which sequence is increasing and which is decreasing?

View Solution

Step 1 — AM-GM: $a_{n+1} = \frac{a_n + b_n}{2} \ge \sqrt{a_n b_n} = b_{n+1}$ , so $a_{n+1} \ge b_{n+1}$ for all $n$ .

Step 2 — $\{a_n\}$ is Decreasing: Since $b_n \le a_n$ :

a_{n+1} = \frac{a_n + b_n}{2} \le \frac{a_n + a_n}{2} = a_n

Step 3 — $\{b_n\}$ is Increasing: Since $a_n \ge b_n$ :

b_{n+1} = \sqrt{a_n b_n} \ge \sqrt{b_n \cdot b_n} = b_n

Step 4 — Bounded: $\{a_n\}$ is decreasing, bounded below by $b_1$ . $\{b_n\}$ is increasing, bounded above by $a_1$ .

Step 5 — Convergence: By the Monotone Convergence Theorem, both converge. Let $L = \lim a_n$ , $M = \lim b_n$ :

L = \frac{L + M}{2} \implies L = M

Both converge to the arithmetic-geometric mean $\operatorname{AGM}(a, b)$ .

15 min Putnam 1947, Problem A5

VI. Monotonicity Across Mathematics

The Erdős–Szekeres Theorem reveals that monotonicity is not merely a calculus concept — it is a universal structural phenomenon.

The Erdős–Szekeres Theorem

Problem Hard

Prove that any sequence of $n^2 + 1$ distinct real numbers contains a monotone subsequence of length $n + 1$ .

View Solution

The Pigeonhole Synthesis:

For each element $a_i$ , assign the pair $(r_i, d_i)$ where:

$r_i$ = length of the longest increasing subsequence ending at $a_i$ .
$d_i$ = length of the longest decreasing subsequence ending at $a_i$ .

Assume no monotone subsequence of length $n+1$ exists. Then $r_i \le n$ and $d_i \le n$ for all $i$ , giving at most $n^2$ distinct pairs.

Since there are $n^2 + 1$ elements, by the Pigeonhole Principle, two indices $i < j$ share the same pair.

Contradiction:

If $a_i < a_j$ : $a_j$ extends the increasing subsequence ending at $a_i$ , so $r_j \ge r_i + 1$ . Contradicts $r_i = r_j$ .
If $a_i > a_j$ : $a_j$ extends the decreasing subsequence, so $d_j \ge d_i + 1$ . Contradicts $d_i = d_j$ .

Since elements are distinct, one case must hold. $\blacksquare$

15 min Competition Folklore

Cross-Connections

Number Theory: Multiplicative Rigidity

Erdős’s Theorem: If a multiplicative function $f: \mathbb{N} \to \mathbb{R}^+$ is monotonically increasing, it must be of the form $f(n) = n^{\alpha}$ for some $\alpha \ge 0$ . The algebraic structure (multiplicativity) combined with order structure (monotonicity) forces the function into an extremely narrow form — a beautiful example of algebraic rigidity.

Differential Equations: Picard Iteration

In solving $y' = F(x, y)$ , if $F$ is monotone in $y$ , we can construct upper and lower solutions that squeeze the actual solution — the Monotone Iterative Method. The convergence relies on the Monotone Convergence Theorem, connecting pure analysis to computational trajectories.

Selected Problems

Problem Hard

(Putnam 1968 B6 variant): Let $f: [0,1] \to \mathbb{R}$ be continuous. Suppose that for every $x \in [0,1]$ , there exists $\delta > 0$ such that $f$ is monotone on $(x - \delta, x + \delta) \cap [0,1]$ . Prove that $f$ is monotone on $[0,1]$ .

Hint

Use compactness and the local-to-global principle.

Problem Hard

(ISI B.Math Style): Let $f: \mathbb{R} \to \mathbb{R}$ be strictly increasing on $\mathbb{Q}$ and continuous on $\mathbb{R}$ . Prove that $f$ is strictly increasing on $\mathbb{R}$ .

Hint

Use the density of

\mathbb{Q}

\mathbb{R}

: every real number is the limit of a sequence of rationals.

Problem Medium

(Erdős–Szekeres Extension): Construct a permutation of $\{1, 2, \ldots, n^2\}$ that has no increasing or decreasing subsequence of length $n + 1$ .

Hint

Glue together

n

decreasing blocks of length

n

. For example, with

n=3

(3, 2, 1, 6, 5, 4, 9, 8, 7)

Problem Advanced

(Functional Monotonicity): Find all $f: \mathbb{N} \to \mathbb{R}^+$ such that $f(mn) = f(m)f(n)$ for all $m, n$ and $f$ is monotonically increasing.

Hint

Erdős’s theorem forces

f(n) = n^{\alpha}

. Start by showing

f(1) = 1

and

f(p^k) = f(p)^k

Problem Hard

(Discontinuous Derivative): Prove that if $f$ is differentiable on $[a, b]$ and $f'$ is monotonic, then $f'$ must be continuous.

Hint

Darboux’s Theorem guarantees

f'

has the IVT property. Show that a monotone function with the IVT property cannot have jump discontinuities.

Problem Medium

(Sequence Limits): Evaluate $\displaystyle\lim_{n \to \infty} \frac{1}{n} \sum_{k=1}^{n} \left\lfloor \frac{2n}{k} \right\rfloor \cdot \frac{1}{n}$ .

Hint

Interpret as a Riemann sum and use the monotonicity of the integrand.

Problem Hard

(Functional Equation): Find all continuous $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x+1) = f(x) + 1$ and $f(f(x)) = x$ for all $x$ .

Hint

f

must be monotonic (why?). Consider what

f(f(x)) = x

and

f(x+1) = f(x)+1

together force.

Problem Medium

(Inverse Integration): Let $f: [0,a] \to [0,b]$ be strictly increasing, continuous, with $f(0) = 0$ . Prove:

\int_0^a f(x) \, dx + \int_0^b f^{-1}(y) \, dy = ab

Geometric Hint

Draw the rectangle

[0,a] \times [0,b]

. The graph of

f

splits it into two regions — one is

\int f

, the other is

\int f^{-1}

Challenge Problem

Problem Advanced

The Devil’s Slide

Let $c: [0,1] \to [0,1]$ be the Cantor function (Devil’s Staircase). It is continuous, non-decreasing, surjective, with $c(0) = 0$ , $c(1) = 1$ , and $c'(x) = 0$ almost everywhere — yet it is not constant.

Part (a): The Cantor function has plateaus (it is not strictly increasing). Construct a modification $g: [0,1] \to [0,1]$ that is:

Strictly increasing,
Singular ( $g'(x) = 0$ almost everywhere),
A homeomorphism of $[0,1]$ to itself.

Part (b): Prove that such a $g$ maps a set of Lebesgue measure 0 to a set of Lebesgue measure 1.

Construction Hint

Consider

g(x) = \frac{1}{2}(c(x) + x)

. Adding the identity “tilts” the staircase into a strictly increasing function while preserving singularity.