The sequence $f_k(x) =k(1-kx), x \in [0,1/k], f_k(x) = 0, x \in [1/k,1] $ converges a.e. to $f(x) = 0$ in $[0,1]$ so $|f_k(x)|\ge 0 \text{ a.e. in } E :=[0,1] $
The hypotheses of the dominated convergence thm thus hold and I should find that $\displaystyle\lim_{k \to \infty}\int_{E}|f_k(x)-f(x)|dx \to 0 $ but in this case $$\displaystyle\lim_{k \to \infty}\int_{E}|f_k(x)-f(x)|dx= \lim_{k \to \infty} \int_{[0,1/k]}|k(1-kx)|dx=1/2 \nrightarrow 0 $$
It's $|f_k(x)| \le 0$ :p
Now that I think about it $[0,1/k]$ only becomes a set of measure 0 as $k \to \infty$
ok yeah there's something fishy: the pointwise limit is indeed weird: $$f_k(x) \to f(x):= \begin{aligned}
0, x \in (0,1]\\ +\infty, x = 0 \end{aligned} $$
Ok maybe I cannot bound my sequence of functions with 0
I have a basic question, which I'm not so sure about. Is every compact set in $\mathbb R$ of the form $[a,b]$ where $-\infty<a<b<\infty$? I know if it is of this form, it is compact. But I'm kind of asking for the converse.
No, not even if you extend to finite union of segments. A set in R is compact iff it is closed and bounded So for example $\{\frac{1}{n}\mid n \in \mathbb{N}\} \cup \{0\}$ is compact
Hi, guys. Basic question on smooth manifolds: I have an ambient manifold and a submanifold. Consider then a function and vector field on the submanifold. I can extend them to the ambient. I want to apply the vector field on the function (at a certain point of the submanifold). This will give me a number on this point (a function on the submanifold). Using the extensions instead but restricted to that point (submanifold) gives me the same number (function)?
@Derso yes, this is true, essentially because the map $T_pN\rightarrow T_pM$ induced by the inclusion $N\subseteq M$ is given by restricting germs along that inclusion
@psie the compact sets in $\mathbb{R}$ are precisely the complements of unions of collections of open intervals that contain both an interval of the form $(-\infty,a)$ and an interval of the form $(b,\infty)$
ok so you have a sequence of (measurable) functions $(f_k(x))_k$, suppose that $\lim f_k(x)=f(x)$. If $|f_k(x)|\le g(x)$ a.e. for all $k$ and $g\in L^1$, then you can swap $\lim$ and $\int$
@Jakobian I was joking obv, as I said yesterday, uniform convergence requires the most efforts to prove so I often can't do it, therefore I get frustrated :p
@Jakobian yeah I agree solving problems that tingle one's brain is motivating and rewarding
yeah, it's essentially obvious, which is why it puzzles me that Lurie presents such a long-winded proof
you just have to observe that the identity $1_{\mathcal{D}}=ps$ is a unit transformation for the adjunction, i.e. that $p$ induces an equivalence $\mathrm{map}_{\mathcal{C}}(sd,c)\rightarrow\mathrm{map}_{\mathcal{D}}(d,pc)$ for all $c,d$, but that's to say that $sd$ is a $p$-initial vertex and, for a cartesian fibration, being relative initial is the same as being initial in the fiber, as shown somewhere in chapter 4
Of course, you also need to recognize that the baseline is often taken to be the period from around 1950-1990, as this is when the tools for studying climate change, and the recognition that it was going on, developed.
Things had already started getting warmer, so it makes sense that previous history looks"cold".
![[After setting your car on fire] Listen, your car's temperature has changed before.](https://imgs.xkcd.com/comics/earth_temperature_timeline.png "[After setting your car on fire] Listen, your car's temperature has changed before.")
