A Simple Example of a Function on $[0, 1]$

Question

I am looking for a concrete example of a continuous function $f$ on the unit interval $[0, 1]$ such that $0\leq f(x) <1$ for each $x\in [0, 1]$, but $\int_0^1 f(x) dx = 1$.

Any help is greatly appreciated.

Answer 1

There is no such function. EDIT: as Greg Martin points out, my original answer was a bit silly. One should just say that any continuous function on a closed interval achieves its maximum. Since $f(x)<1$, this maximum $M$ is also less than $1$, and

$$\int_0^1 f(x)dx < \int_0^1 Mdx = M < 1$$

---

(original answer:)

Since $f(x) < 1$ for any $x$, $1 - f(x)$ is a positive function for $x\in[0,1]$. Any continuous function on a closed interval has a minimum, so for some number $\epsilon$ we have $1 - f(x) > \epsilon$. Rearrange to get $f(x) < 1 - \epsilon$.

Then the integral is similarly bounded: $$\int_0^1 f(x)dx < \int_0^1 (1-\epsilon) dx = 1 - \epsilon$$

Answer 2

No such function exists.

Suppose by way of contradiction that such a function existed. Let $c\in (0,1).$ Then $0\leq f(c)<1.$ Furthermore, $f$ is continuous, which means $\exists\ 0<\delta<c$ such that

$$ \left\vert x-c \right\vert < \delta \implies \left\vert f(x)-f(c) \right\vert< \frac{1-f(c)}{2}=\varepsilon.$$

But this implies that $\displaystyle\int_0^1 f(x) dx \leq 1 - \left(\frac{1-f(c)}{2}\right)2\delta<1.$