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The maps $F_K$ are certainly upper semicontinuous, because, by the regularity property of the Lebesgue measure, for every $x\in M$ and for $\epsilon>$ there exists $\delta>0$ such that the uniform $\d …

As to: any way of proving [the derivative of the delta function] is not the difference of two positive distributions: indeed, $\delta’$ is not even majorized by a positive distribution. Here’s a dir …

Another possibility is an approximation of the identity via convolution; by linearity it smoothes the $1_{\Delta_i}$, and of course $\|f*\phi_\epsilon\|_\infty\le \|f \|_\infty$, because the convoluti …

Here is a statement in a slightly more general setting (since it is for free). Let $X$ be a length metric space, $Y$ a metric space, $f:X\to Y$ continuous, $(A_i)_{i\in \mathbb N}$ a countable cove …

As I see it, it would be preferable a simple term not referring to more advanced characterisations of more general objects (“Finite homology subset”? “Semialgebraic subset”? “Submanifold with bounda …

Take $(E,\mathcal E)=(\Omega,\mathcal G)= \mathbb R$ with the Borel $\sigma$-algebra, and let $C$ be a subset of the diagonal $\Delta\subset \mathbb R^2$. If for a linear combination $g$ of rectangles …

One reasonable explanation, to me, is that the above is nothing but the convexity inequality $f\big(t u+(1-t)v\big)\le tf(u)+(1-t)f(v)$ for the function $-\log$, changing the names of the variables a …

Let $G$ be a free group on $p>1$ generators and $X=G$. It is a locally compact complete metric space with its invariant distance defined by length of words. The invariant measure given by cardinality …

The answer is no, for quite a trivial reason: $\mathcal{L}$ may have not enough pairs $A\subset B$, and not enough monotone increasing sequences $(A_n)_n$, to make $(a)$ and $(b)$ meaningful. For inst …

The usual equi-continuity $L^1$ is $$\sup_n\|f_n-\tau_h f_n \|_1=o(1)\ \qquad \text{as } h\to0$$ (Here $(\tau_hf)(x)=f(x+h)$; the $f_n$ are to be zero-extended to $\mathbb R$ so that $\tau_h f_n$ is …

Here are some details on Sam Forster’s construction . To make the computation simpler I’d take powers of $4$ instead, i.e. define $g:= \sum_{k=1}^\infty \frac{f_k}{4^k},$ where $f_n$ and $C_n$ have b …

There are no such $A,B$. If $A\subset[0,1]$ is a Lebesgue measurable set of positive measure, the set $\mathbb Q_+ A:=\{qa: q\in\mathbb Q_+,\, a\in A\}$ has full measure in $\mathbb R_+$, for it has a …

Recall: 1. Given $f_n\in C(X,Y)$ point-wise convergent to $f$, the above set $E$ always contains the discontinuity set of $f$ (starting from any $x_n\to x$ with $ f(x_n)\not\to f(x)$ one has for a sub …

An elementary non-existence proof may be of interest. Let $f:[a,b]\to\mathbb{R}$ an increasing, continuous, and not absolutely continuous function: I claim there exists a point $c\in[a,b]$ where the D …

I'll give a picture, though not entering into all details. Writing $g(\epsilon,x):=g_\epsilon(x)$ and $I:=]-\delta,\delta[$, a natural assumption is that $g:I\times K\to\mathbb R$ is $C^1$, and $\mu$ …