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Continuous $f_n \to 0$ pointwise on $[0,1]$ with $\int_0^1 f_n = 1$ for every $n$

Statement refuted

False claim: if $(f_n)$ is a sequence of Riemann integrable functions on $[0,1]$ converging pointwise to $f$ , and $f$ is integrable, then $\int_0^1 f_n \to \int_0^1 f$ .

For $n \in \mathbb{N}$ write $c_n := \iota(n+1) \ge 1$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field, Canonical naturals are positive and strictly increasing) and define the tent $f_n : [0,1] \to \mathbb{R}$ by

$f_n(x) \;:=\; \begin{cases} 4c_n^{2}\,x & 0 \le x \le \tfrac{1}{2c_n}, \\[3pt] 4c_n^{2}\bigl(\tfrac{1}{c_n} - x\bigr) & \tfrac{1}{2c_n} \le x \le \tfrac{1}{c_n}, \\[3pt] 0 & \tfrac{1}{c_n} \le x \le 1 . \end{cases}$

Each $f_n$ is continuous on $[0,1]$ , hence integrable, with

$\int_0^1 f_n \;=\; 1 \qquad \text{for every } n ,$

while $f_n(x) \to 0$ for every $x \in [0,1]$ . So the pointwise limit is the zero function, whose integral is $0$ , and the integrals do not converge to it.

The heights are unbounded: $f_n$ attains the value $2c_n$ at $x = 1/(2c_n)$ , and $2c_n \to \infty$ . That is what the example refutes and what it does not: it refutes the interchange for pointwise convergence, and it says nothing whatever about sequences that are uniformly bounded, for which no theorem is stated on this page in any direction.

Facts & Assumptions

Given: For $n \in \mathbb{N}$ , $c_n = \iota(n+1)$ and the function $f_n$ above; a point $x \in [0,1]$ and a real $\varepsilon > 0$ .

[L1]

$\iota(n+1) \ge 1 > 0$ , $\iota$ is nondecreasing on $\mathbb{N}$ , and for every real $\eta>0$ there is a natural $N$ with $1/\iota(N+1) < \eta$ (The canonical natural $\iota(n) = n \cdot 1_F$ of a field, Canonical naturals are positive and strictly increasing, Every complete ordered field is Archimedean, For every $\varepsilon > 0$ in a complete ordered field there is a natural $n \ge 1$ with $1/n < \varepsilon$ ).

[L2]

Every polynomial function is continuous, and a function agreeing with continuous functions on the pieces of a finite subdivision of $[0,1]$ , with matching values at the shared endpoints, is continuous on $[0,1]$ (Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, claim 5, Continuity of $f : A \to \mathbb{R}$ at a point of $A$ and on $A$ : the $\varepsilon$ - $\delta$ condition, its agreement with $\lim_{x \to c} f(x) = f(c)$ at a limit point, and continuity at an isolated point).

[L3]

A continuous function on a closed bounded interval with distinct endpoints is integrable there (A continuous function on $[a,b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, The lower and upper Darboux integrals of a bounded $f$ on $[a,b]$ as $\sup_P L(f,P)$ and $\inf_P U(f,P)$ , Darboux integrability as their equality, and the notation $\int_a^b f$ ).

[L4]

Additivity over adjacent subintervals (For $a<c<b$ : $f$ is integrable on $[a,b]$ if and only if it is integrable on $[a,c]$ and on $[c,b]$ , and then $\int_a^b f = \int_a^c f + \int_c^b f$ ; with the oriented form for arbitrary $a,b,c$ , The integral with oriented limits: $\int_a^a f := 0$ and $\int_b^a f := -\int_a^b f$ ).

[L5]

If $H$ is differentiable at every point of $[p,q]$ with $H'$ integrable there, then $\int_p^q H' = H(q)-H(p)$ ; a continuous function on an interval has a primitive (The second fundamental theorem: if $G$ is differentiable on $[a,b]$ with $G' = f$ and $f$ is integrable, then $\int_a^b f = G(b)-G(a)$ , Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_a^b f = G(b)-G(a)$ for any primitive $G$ ).

[L6]

For $n \ge 1$ the map $x \mapsto x^{n}$ has derivative $\iota(n)x^{\,n-1}$ ; sums and scalar multiples of differentiable functions are differentiable with the corresponding derivatives; and $\int_p^q c = c(q-p)$ for a constant (For a natural $n \ge 1$ the function $x \mapsto x^{n}$ is differentiable everywhere with derivative $\iota(n)\,x^{\,n-1}$ ; for $n = 0$ it is the constant $1$ , with derivative $0$ ; for a natural $n \ge 1$ the function $x \mapsto x^{-n}$ is differentiable at every $x \ne 0$ with derivative $-\iota(n)\,x^{-n-1}$ ; consequently every polynomial function is differentiable at every real, with the derivative computed term by term, Sums, scalar multiples, products and quotients: $(f+g)'(c) = f'(c) + g'(c)$ , $(\alpha f)'(c) = \alpha f'(c)$ , $(fg)'(c) = f'(c)g(c) + f(c)g'(c)$ , and $(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2}$ when $g(c) \ne 0$ , If $m \le f \le M$ on $[a,b]$ then $m(b-a) \le L(f,P) \le \underline{\int_a^b} f \le \overline{\int_a^b} f \le U(f,P) \le M(b-a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_a^b c = c(b-a)$ , The derivative $f'(c) = \lim_{x \to c} \frac{f(x) - f(c)}{x - c}$ of $f : A \to \mathbb{R}$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Integer powers $a^m$ ).

[L7]

A sequence of reals converges to $L$ when for every real $\varepsilon>0$ there is $N$ with $|a_n - L| < \varepsilon$ for all $n \ge N$ (Limits and Cauchy sequences of reals, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L8]

Ordered-field arithmetic and linearity of the integral: multiplying inequalities by positive reals, transitivity, and $\int_p^q(\lambda u) = \lambda\int_p^q u$ (Ordered field, Complete ordered field (least-upper-bound property), Integrable functions on $[a,b]$ form a set closed under sums and scalar multiples, and $\int_a^b(\lambda f+\mu g) = \lambda\int_a^b f + \mu\int_a^b g$ , Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length).

Counterexample

technique · direct

1.1

$c_n \ge 1$ by [L1], so $0 < 1/(2c_n) < 1/c_n \le 1$ and the three pieces of the definition subdivide $[0,1]$ .

givenL1

2.1

The three formulas agree at the shared endpoints: at $x = 1/(2c_n)$ both give $4c_n^{2}/(2c_n) = 2c_n$ , and at $x = 1/c_n$ the second gives $0$ , which is the third. So $f_n$ is a well-defined function and is continuous on $[0,1]$ by [L2], hence integrable by [L3].

step 1.1L2L3

2.2

On $[1/c_n, 1]$ the function $f_n$ is constantly $0$ , so $\int_{1/c_n}^{1} f_n = 0$ by [L6]; when $1/c_n = 1$ this piece is degenerate and the integral is $0$ by [L4].

step 1.1L4L6

2.3

Pointwise convergence to $0$ . At $x = 0$ every $f_n(0) = 0$ . For $x > 0$ , [L1] gives $N$ with $1/\iota(N+1) < x$ , and for $n \ge N$ one has $c_n = \iota(n+1) \ge \iota(N+1)$ , hence $1/c_n \le 1/\iota(N+1) < x$ , so $x$ lies in the third piece and $f_n(x) = 0$ .

step 1.1givenL1L8

3.1

On $[0, 1/(2c_n)]$ the function $H(x) := 2c_n^{2}x^{2}$ has $H' (x)= 4c_n^{2}x = f_n(x)$ by [L6], so $\int_0^{1/(2c_n)} f_n = H(1/(2c_n)) - H(0) = 2c_n^{2}/(4c_n^{2}) = 1/2$ by [L5].

step 2.1L5L6

3.2

On $[1/(2c_n), 1/c_n]$ the function $H_2(x) := 4c_n^{2}\bigl(x/c_n - x^{2}/2\bigr)$ has $H_2'(x) = 4c_n^{2}(1/c_n - x) = f_n(x)$ by [L6], and $H_2(1/c_n) = 4c_n^{2}\bigl(1/c_n^{2} - 1/(2c_n^{2})\bigr) = 2$ while $H_2(1/(2c_n)) = 4c_n^{2}\bigl(1/(2c_n^{2}) - 1/(8c_n^{2})\bigr) = 3/2$ ; so $\int_{1/(2c_n)}^{1/c_n} f_n = 2 - 3/2 = 1/2$ by [L5].

step 2.1L5L6

3.3

Hence $|f_n(x) - 0| = 0 < \varepsilon$ for all $n \ge N$ , so $f_n(x) \to 0$ for every $x \in [0,1]$ by [L7].

step 2.3L7

4.1

By [L4] applied twice, $\int_0^1 f_n = 1/2 + 1/2 + 0 = 1$ for every $n$ .

step 3.1step 3.2step 2.2L4

5.1

The pointwise limit is the zero function, which is integrable with integral $0$ by [L6], while $\int_0^1 f_n = 1$ for every $n$ by step 4.1; so the integrals do not converge to the integral of the limit and the claim is false.

step 4.1step 3.3L6L7∎

Remarks

What this refutes, stated exactly. It refutes the interchange of a limit with an integral under pointwise convergence alone, even when every $f_n$ is continuous and the limit function is as regular as possible. It does not refute, and does not address, any statement about uniformly convergent sequences or about uniformly bounded ones; no such statement is proved on this page, and none is contradicted here.
Unboundedness of the heights is essential to the construction and is stated as a feature, not hidden. $\sup_{[0,1]} f_n = 2c_n$ grows without bound, and the mass $1$ escapes into a spike of shrinking width. A reader who wants a theorem in this direction should note that the hypothesis to look for is a bound on the whole sequence, and that whichever theorem supplies it is not on this page.
The integral of the limit exists here. The failure is not that the limit function is non-integrable — it is the zero function — but that the numbers $\int_0^1 f_n$ simply do not converge to $\int_0^1 0$ . A separate failure, in which the pointwise limit of integrable functions is not integrable at all, is recorded as a false statement on the companion page of The Riemann Integral.

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Sources

Riemann integral (Wikipedia) (standard reference, not scraped)
Uniform convergence (Wikipedia) (standard reference, not scraped)
UC Davis MATH 125B, Chapter 1 notes (standard reference, not scraped)

Continuous fn→0f_n \to 0fn​→0 pointwise on [0,1][0,1][0,1] with ∫01fn=1\int_0^1 f_n = 1∫01​fn​=1 for every nnn