If I have $h(x)=1_A(x) \, f(x) = \begin{cases}f(x) & \text{if } x \in A \\ 0 & \text{if } x \not\in A\end{cases}$, where $A$ is an arbitrary set, then why it's equal to $f(x)$ almost everywhere?
Is there equality between an infinity norm of a function and its primitive?
2026-03-22 04:44:58.1774154698
limit of characteristic function and relation of primitive
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It isn't in general. For example, with $f(x)\equiv 1$ and $A=\emptyset$ we have $1_A(x) \, f(x) \equiv 0 \neq f(x) \equiv 1$ everywhere.
No, there isn't. For example, $f_n(x) = \sin nx$ where $n=1,2,3,\ldots$ all have $\|f_n\| = 1$ but $\|f_n'\|=n$. Thus there is no such equality.