I am running into some troubles with Lipschitz continuous functions. Suppose I have some one-dimensional Lipschitz continuous function $f : \mathbb{R} \to \mathbb{R}$. How do I prove that its derivative exists almost everywhere, with respect to the Lebesgue measure? I found on other places on the internet that any Lipschitz continuous function is absolutely continuous, […]

Let $f_n$ be a sequence of Lebesgue measurable functions on $R^d$. Suppose you have an estimate of the form $\int_{R^d}\left|f_n\right|\le c_n$ where $c_n \downarrow 0$. Can you conclude that $f_n\to 0$ a.e.? If not, what additional conditions on ${c_n}$ would guarantee this? My attempt: I think we cannot conclude that $f_n\to 0$ a.e. For example […]

Let $X_n$ be a sequence of independent random variables, with $\mathbb E(X_n)=0$ and $|X_n|<K, \forall n,\omega$, then: P$(\sup_{n\in\mathbb N}|\sum_{k=1}^{n}X_k|<\infty)>0 \iff$ P$(\sum_{k=1}^{\infty}X_k$ exists in $\mathbb R)=1$ This can be also found in David Williams, probability with martingales (Remark in p.113). But this is not proven there. New question deleted. It can be found here.

Let $ f \in L^1[0,1]$. Assume that there is a constant C, with $0 < C < 1$, such that for every measurable set $A \subset [0,1] $ with $m(A)=C$, we have $ \int_{A} f dm = 0 $. Prove that $f = 0$ almost everywhere. I tried to do my contradiction but I could […]

I am having some conceptual difficulties with almost everywhere (a.e.) convergence versus convergence in measure. Let $f_{n} : X \to Y$. In my mind, a sequence of measurable functions $\{ f_{n} \}$ converges a.e. to the function $f$ if $f_{n} \to f$ everywhere on $X$ except for maybe on some set of measure zero. However, […]

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