Measurable function but not in space $L_{\infty}(0,1)$

Question

Can someone give an example of measurable (by Lebesgue) function $f:(0,1) \to \Bbb R$ which will not be in space $L_{\infty}(0,1)$?

Answer 1

You can take any unbounded but continuous function.

Continuous functions are measurable for sure, but if it's unbounded, it's not in $L^\infty$.

A concrete example would be: $f(x) = \frac{1}{1-x}$, which diverges as $x \to 1$.

Or take step functions that increase, something like $$g(x) = 1 \ \text{for} \ x \in (0,0.5)$$ $$g(x) = 2 \ \text{for} \ x \in [1/2,3/4)$$ $$g(x) = 4 \ \text{for} \ x \in [3/4,7/8)$$ $$g(x) = 8 \ \text{for} \ x \in [7/8,15/16)$$ $$...$$ and so on...

Answer 2

$$f(x) := \ln\left( \frac{1}{x} \right)$$ is an example. It is measurable because it is a composition of two functions that are continuous on $(0,1)$. This is a nice example to keep around because $f \in L^p(0,1)$ for all finite $p$, but not $p = \infty$.

Edit: To see why $f \notin L^\infty$, fix $\alpha > 0$. Then $|f(x)| = f(x) > \alpha$ if and only if $$\ln\left( \frac{1}{x} \right) > \alpha \Longleftrightarrow \frac{1}{x} > e^\alpha \Longleftrightarrow x < \frac{1}{e^\alpha} . $$

So $m(\{ x \in (0,1) : |f(x) | > \alpha \}) = m((0, 1/e^\alpha)) = 1/e^\alpha$ and since $\alpha > 0$ was arbitrary, the set $\{ x \in (0,1) : |f(x) | > \alpha \}$ is always positive, which implies that $$ \| f \|_{\infty} = \inf \{ \alpha \in \mathbb{R} : m(\{ x \in (0,1) : |f(x) | > \alpha \}) = 0\} = \inf \emptyset =\infty.$$