Articles of almost everywhere

Let $f$ be measurable and $a,b\in\mathbb{R}$ with $\frac{1}{\lambda(M)}\int_Mf\ d\lambda \in $ Show that: $f(x) \in $ almost everywhere.

Assignment: Let $f$ be Lebesgue – measurable and $a,b \in \mathbb{R}$ with the property: $$\frac{1}{\lambda(M)} \cdot \int_Mf\ d\lambda \in [a,b]$$ for all Lebesgue – measurable sets $M \subset \mathbb{R}^n$ with $0 < \lambda(M) < \infty$. Show that: $f(x) \in [a,b]$ almost everywhere. So I need to show that the set $N$ defined as $N:= \{x\in\mathbb{R}^n; \ f(x) […]

Doubt about eqivalence .

Regarding the second line of the first proof by @JDL in Altenate definitions of almost sure convergence Are the following equivalent? $P(\omega:\exists n\in\mathbf{N}, \forall m\in\mathbf{N}, \exists i>m \,\,\, \text{s.t.} \,\, |X_i(\omega) – X(\omega)| < 1/n)$ AND $P(\omega:\forall n\in\mathbf{N}, \exists m\in\mathbf{N},\ \text{s.t.}\ \forall i>m \,\, |X_i(\omega) – X(\omega)| \ge 1/n)$

version of the dominated convergence theorem where the almost-everywhere convergence is used

Let $f\in \mathcal{L}^0(S,\mathcal{S},\mu)$ be a function State and prove a version of the dominated convergence theorem where the almost-everywhere convergence is used. Is it necessary for all $\{f_n \}_{n∈\mathbb N}$ to be dominated by $g$ for all $x\in S$, or only almost everywhere? I don’t even have a direction. What do I need to show? […]

Almost sure identity $F^{-1}(F(X))=X$ where $F$ is the CDF of $X$

Let $F$ be the distribution function of a random variable $X$. If $F$ is continuous, then it holds that $F^{-1}(F(X))=X$ almost surely, where $F^{-1}$ denotes the generalized inverse of $F$. My question is now: Why does this result not hold for a discontinuous function $F$? Generally it holds that $F^{-1}(F(x))\leq x$. Assume that $F$ is […]

Lipschitz continuity implies differentiability almost everywhere.

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, […]

Under what condition can converge in $L^1$ imply converge a.e.?

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 […]

Prove P$(\sup_{n\in\mathbb N}|\sum_{k=1}^{n}X_k|<\infty)>0 \iff$ P$(\sum_{k=1}^{\infty}X_k$ exists in $\mathbb R)=1$

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.

If $\int_A f\,dm = 0$ for all $A$ having some fixed measure $C$, then $f = 0$ almost everywhere

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 […]

Almost Everywhere Convergence versus Convergence in Measure

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, […]