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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 …

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 …

You do not need $\sigma$-finiteness of the measure in Fubini theorem, although it is an hypothesis that can be assumed with no loss of generality, in that the support of an integrable function is, of …

Although I agree, let's recall that "being intuitive" is quite a relative matter. In the present case, I find that the Carathéodory's settlement is optimal: maximal effect, minimal effort; maximal gen …

A counterexample is a signed measure on the interval $I:=[-1,1]$ concentrated in the points $\{-1\}$, $\{0\}$, $\{1\}$ with weights respectively $1/2$, $-1$, $1/2$. (Thus $\int_If d\mu= f(1)/2+f(-1)/ …

If $X$ is an infinite dimensional separable Banach space (like $C^0(\Omega)$, for a compact metric space $\Omega$ ), and $\{y _ j \} _ {j\ge 1}$ is a dense sequence in its unit ball, one considers th …

Here's a nice application of measure theory, precisely, of the the theory of orthogonal polynomials, to a classic problem of counting derangements. Problem: How many anagrams with no fixed letters of …

To summarize the situation. Let $(X,\mathcal{F})$ a measurable space. The space $M(X,\mathcal{F})$ of all real-valued signed measures on $(X,\mathcal{F})$ is a Banach space wrto the total variation no …

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 …

Let us consider more closely the question about space-filling curves. The Peano curve and the Hilbert curve, and several other variations of them, have parametrizations $[0,1]\to[0,1]^2$ that actuall …

I think it is true (even in the more general setting of second countable topological spaces). Let $\{A_n\}_{n\in\mathbb{N}}$ be a countable basis of the topology of $X$. For ${n\in\mathbb{N}}$ let $\ …

There are Hausdorff spaces all of whose real-valued functions are constant. A classical example (of a countable space) was given by Uryshon (Über die Mächtigkeit der zusammenhängenden Mengen, Math.An …

This is the (a version of the) Vitali set, which is not Lebesgue measurable. A quick reason is: $\mathbb{R}= (v_0\mathbb{Q})\oplus_\mathbb{Q} V$ shows that $\mathbb{R}$ is a countable union of tran …

A useful density lemma is the following. Let $(X,\mathcal{A}, \mu)$ be a measure space and let $\Gamma\subset\mathcal{P}(X)$ a ring of sets of finite measure that generates the $\sigma$-alg …

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 …