Solution to Exercises 1, 7 in Tutorial Note 10

There are already too much material, I will not give the solution to all problems in the last tutorial note (which I did in class for tutorial note 1-9). Final exam's problem are more important.

Tutorial Note 10 (Lebesgue integration and convergence theorems)

Tutorial Note 9 (Lebesgue measurable functions)

Remark. We have a minor correction in page 4:

$m_*$ is not subadditivity

Tutorial Note 8 (Lebesgue inner, outer measures and Lebesgue measure)

Tutorial Note 7 (set operations with functions; topology on $\R$: openness, closedness and compactness)

Midterm Review

You can download it here. We will just focus on problems involving $\limsup$ and $\liminf$.

Tutorial note 6 (consequence of uniform convergence)

Tutorial Note 5 (sequence and series of functions; uniform and nonuniform convergence)

Tutorial note 4 (limit inferior and superior)

Remark. Whether we use $\limsn,\limin$ or $\dis\limsup_{n\to\infty},\liminf_{n\to\infty}$ is just a matter of taste. In the tutorial notes I will use $\lims,\limi$ (partly because this is the first notation I see when I learn this concept).

Remark. Note that Theorem 6 is a direct consequence of the definition of $\lims$ and $\limi$, we don't need to seek help from $M_k$/$m_k$ Theorem. To see this, suppose that given $r\in \R$ and $\{a_n\}$ a sequence such that \[
\lims a_n<r,
\] then can there be infinitely $a_n$'s such that $a_n\in [r,\infty)$? NO! Indeed, if this happens, then we can extract a subsequence which diverges to $+\infty$ or converges to $\ell\ge r$, either case contradicts to $\lims a_n<r$.

Remark. Theorem 6 actually enables us to characterizes $\lims a_n$ for any bounded sequence $\{a_n\}$ in the following way: \[
\lims a_n=\min \underbrace{\{r\in \R:\forall \epsilon>0, a_n<r+\epsilon\quad\text{for all $n$ big enough}\}}_{:=\dis \mathcal S}.
\tag*{(*)}
\] Indeed, $r=\lims a_n$ is really such a number by theorem 6, i.e., $\lims a_n\in \mathcal S$. To see this, for any $\epsilon>0$, we have \[
\lims a_n<\lims a_n+\epsilon,
\] therefore by Theorem 6 there is an $N$ such that $n>N$ $\implies$ $a_n<\lims a_n+\epsilon$, thus $\lims a_n\in \mathcal S$. On the other hand, suppose $r\in \mathcal S$, then for any $\varepsilon>0$, there is an $N$ such that \[n>N\implies a_n<r+\varepsilon,\] we take $\lims $ on both sides to obtain $\lims a_n\leq r+\varepsilon$. As this is true for each $\varepsilon>0$, we take $\varepsilon\to 0^+$ to conclude $\lims a_n\leq r$, so $\lims a_n = \min \mathcal S$. Therefore we have another definition of $\lims$ as in $(*)$.

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Tutorial note 3 (inverse and implicit function theorem)

Tutorial note 2 (differentiability and $C^1$ theorem)

Amendment of tutorial note 2:
Two changes have been made:

Remarks on Linear Algebra

Tutorial note 1 (review of linear algebra and multivariable calculus)