# Tag Info

Accepted

• 208k
Accepted

### For measurable $f: \mathbb{R} \rightarrow \mathbb{R}$ prove $f(x)$ and $\frac{1}{f(1/x)}$ cannot both be Lebesgue integrable.

This is a cute problem. Suppose both $$\int_{\mathbb R}|f(x)|dx<\infty\text{ and }\int_{\mathbb R}\frac{1}{|f(\frac 1 x)|}dx<\infty$$ hold. Note that this implies $f(x)\ne0$ for almost all $x$....
• 24.4k
Accepted

### Borel Measurable Function but not Lebesgue Measurable

The map $f$ is continuous, so the preimage of a Borel subset of $\mathbb{R}^2$ will be a Borel subset of $\mathbb{R}$. Thus $f$ is Borel to Borel measurable. Let $N\subseteq[0,1]$ be a Lebesgue ...
• 10.7k
Accepted

### Can the supremum of an uncountable family of measures be replaced by the supremum over a countable subfamily?

This is not true in general. For instance, if $X$ is uncountable and $\mathcal{A}$ contains all the singletons (say, $X=\mathbb{R}$ and $\mathcal{A}$ is the Borel sets) then the supremum of the delta-...
• 333k

### Function that fails to be differentiable on a set of measure zero.

By outer regularity, there is a nested sequence of open sets $U_n$ so that $E \subseteq U_n$ and $m(U_n) < 2^{-n}$. Let $$f(x) = x + \sum_{n=1}^\infty m(U_n \cap (-\infty, x))$$ To see that $f$ ...
• 453k

### Multivariable function having partial derivatives almost everywhere are necessarily measurable?

Even more general result is true: Theorem. A function $f:\mathbb R^n\to \mathbb R$ is Lebesgue measurable if $f$ is separately continuous almost everwhere. Proof. This theorem can be proved by ...
• 676
Accepted

### Markov kernel in simple random work

I'm not sure what exactly is your question. Markov kernels are a way of specifying Markov processes, extending the concept of transition matrix for (finite-state) Markov chains. More generally, they ...
• 1,985
Accepted

### Suppose $f$ is a measurable function and $f(x) > 0$ for all $x$. Let $g(x) = \frac{1}{f(x)}$. Prove that g is a measurable function.

To show that a function is measurable, it is necessary to show that the preimage of each measurable set is measurable. To do this, it is sufficient to show that the preimage of each interval of the ...
• 30k
Accepted

• 113k
Accepted

### Why isn't the approach of measurable functions (almost) duplicated in the approach to computable functions?

First of all, at a higher level I think it's worth mentioning the extremely well-developed analogy between the hyperarithmetic hierarchy of sets of natural numbers and the Borel hierarchy of sets of ...
• 248k