Function in $L^2$ that doesn't vanishing

Question

Is this statement makes a sense: $f\in L^2(0,1)$ such that $f(x)\ne 0 ,\forall x\in (0,1)$ ?

Answer 1

Not really. If $f$ was such a function, then define $g \in L^2(0,1)$ as

$$g(x) = \begin{cases} 0, \text{if } x = \frac12 \\ f(x), \text{ otherwise}\end{cases}$$

Then $f = g$ almost everywhere so they are equal as elements of $L^2(0,1)$.

However, what does makes sense is to require that $f \ne 0$ almost everywhere, i.e. there exists $N \subseteq [0,1]$ such that $\lambda(N) = 0$ and that $f(x)\ne 0, \forall x \in [0,1]\setminus N$.