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2 votes

Does any measurable function whose integral over any interval is $0$ satisfies $f(x)=0, a.e.$?

In order for $\int_a^b f(x)\; dx$ to be defined (as a Lebesgue integral), we need $f \in L^1(a,b)$. The Lebesgue differentiation theorem then says that for almost every $t$, $$ f(t) = \lim_{\...
Robert Israel's user avatar
2 votes

Le Gall Exercise 1.3

Define, for each $t \in [0,1]$, $(f(t))^{-1}(\mathscr{B}(\mathbb{R}^d)):=\{\{f\in C[0,1]:f(t)\in B\}|B \in \mathscr{B}(\mathbb{R}^d)\}$ the $\sigma$-algebra generated by the coordinate projection $f \...
Snoop's user avatar
  • 15.8k
2 votes

Image of measurable sets under one to one (a.e) functions

That would depend heavily on the measure. In an extreme case, if your measure $\mu$ is concentrated on a singleton then the condition "one-to-one a.e." is satisfied trivially and the ...
Lieven's user avatar
  • 1,853
2 votes

A conjecture about measurable functions

Your conjecture is false. Let $\mathcal H$ be the $\sigma$-algebra of Lebesgue measurable sets in $\mathbb R$. Knowing whether each half-line $(-\infty, x)$ occurs, i.e. whether $X < x$, uniquely ...
Robert Israel's user avatar
1 vote

equivalence for Borel-measurable function

In order for $g$ to be $\mathcal{A}$-measurable, the preimage of every $\mathcal{A}$-measurable set has to be in $\mathcal{A}$. The preimage $f^{-1}$ of any interval $[a,b]$ with $b<0$ is empty. If ...
Gowexx's user avatar
  • 350

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