• General advice: try swapping the orders of limits, sums, integrals, etc.

• Good set / bad set: for measure theory or integrals, try to break a set up into “good” and “bad” subsets, and put bounds on each piece separately.

• Limits:

• Take the $$\limsup$$ or $$\liminf$$, which always exist, and aim for an inequality like \begin{align*} c \leq \liminf a_n \leq \limsup a_n \leq c .\end{align*}
• $$\lim f_n = \limsup f_n = \liminf f_n$$ iff the limit exists, so to show some $$g$$ is a limit, show \begin{align*} \limsup f_n \leq g \leq \liminf f_n \qquad (\implies g = \lim f) .\end{align*}
• A limit does not exist if $$\liminf a_n > \limsup a_n$$.
• Sequences and Series

• If $$f_n$$ has a global maximum (computed using $$f_n'$$ and the first derivative test) $$M_n \to 0$$, then $$f_n \to 0$$ uniformly.
• For a fixed $$x$$, if $$f = \sum f_n$$ converges uniformly on some $$B_r(x)$$ and each $$f_n$$ is continuous at $$x$$, then $$f$$ is also continuous at $$x$$ .
• Equalities

• Split into upper and lower bounds: \begin{align*} a=b \iff a\leq b \text{ and } a\geq b .\end{align*}
• Use an epsilon of room: \begin{align*} \qty{ \forall \epsilon, \,\,a < b + {\varepsilon}} \implies a\leq b .\end{align*}
• Showing something is zero: \begin{align*} \qty{ \forall \epsilon, \,\, {\left\lVert {a} \right\rVert} < {\varepsilon}} \implies a = 0 .\end{align*}
• Continuity / differentiability:

• Show it holds on $$[-M, M]$$ for all $$M$$ to get it to hold on $${\mathbf{R}}$$.
• In higher dimensions: intersect with a ball $$B_R(\mathbf{0})\subset {\mathbf{R}}^n$$ about zero.
• Simplifications:

• To show something for a measurable set, show it for bounded/compact/elementary sets and use approximations in measure.
• To show something for an arbitrary function, try various dense classes of functions: continuous, bounded, compactly supported, simple, indicator functions, etc and use approximations in norm.
• Replace $${\varepsilon}\to 0$$ with an arbitrary countable sequence ($$x_n \to 0$$)
• Note: this is not always helpful, since you now have to predicate over all such sequences.
• Integrals

• Calculus techniques: Taylor series, IVT, MVT, etc.
• Break up $${\mathbf{R}}^n = \left\{{{\left\lvert {x} \right\rvert} \leq 1}\right\} \coprod \left\{{{\left\lvert {x} \right\rvert} > 1}\right\}$$.
• Or break integration region into disjoint annuli: \begin{align*} \int_{\mathbf{R}}f = \sum_{k\geq 0}\int_{2^k}^{2^{k+1}} d .\end{align*}
• For pairs of functions $$f, g$$: break up into $$\left\{{f>g}\right\} {\textstyle\coprod}\left\{{f=g}\right\} {\textstyle\coprod}\left\{{f< g}\right\}$$.
• Tail estimates!
• Most of what works for integrals will work for sums.
• Measure theory:

• Always consider bounded sets, and if $$E$$ is unbounded write $$E = \bigcup_{n\geq 0} \qty{ B_{n}(0) \cap E}$$ and use countable subadditivity or continuity of measure.

• $$F_\sigma$$ sets are Borel, so establish something for Borel sets and use this to extend it to Lebesgue.

• $$s = \inf\left\{{x\in X}\right\} \implies$$ for every $$\varepsilon$$ there is an $$x\in X$$ such that $$x \leq s + \varepsilon$$ or $$x\in [s, s+{\varepsilon}]$$.

• Useful facts about continuous compactly supported ($$C_c^0({\mathbf{R}})$$) functions:

• Uniformly continuous
• Bounded almost everywhere
• Pass to a subsequence!

• Add and subtract a thing. Eg, $${\left\lVert {T_nx_n - Tx} \right\rVert} = {\left\lVert {T_nx_n - Tx_n + Tx_n - Tx} \right\rVert}$$.

• $$(a_k) \in \ell^2({\mathbf{Z}})$$ is much weaker than $$(a_k) \in \ell^1({\mathbf{Z}})$$.

• Littlewood’s principles:

• Measurable sets are almost finite unions of intervals,
• Measurable functions are almost continuous,
• Pointwise convergent sequences of measurable functions are almost uniformly convergent.
• Nesting of $$L^p$$ spaces: let $$p< q$$

• For $$\mu(X) = \infty$$: no general containments.
• For $$\mu(X) < \infty: p < p+1 < \cdots \implies L^p \supseteq L^{p+1} \supseteq \cdots$$. Why? Holder.
• For $$X={\mathbf{Z}}: L^p \subseteq L^{p+1} \subseteq \cdots$$
• Failing to be in $$L^p$$: singularities away from infinity, or long tails.

• Every Borel is $$F_\sigma$$ up to a null set.

• Proving uniform convergence: use the $$M{\hbox{-}}$$test.

• A problem using absolute continuity will often be used to imply bounded variation (which allow using FTC)

• If two functions are in conjugate $$L^p$$ spaces, try Holder.

• $$\mu(X) = {\left\lVert {\operatorname{id}} \right\rVert}_{L^1(X)} = \int_X 1 \,d\mu$$

# The Absolute Essentials

\begin{align*}\sum a_n < \infty \implies a_n \to 0 {\quad \operatorname{and} \quad} \sum_{k=N}^\infty a_n \overset{N\to\infty}\to 0\end{align*}

If $$f_n\to f$$ pointwise and uniformly with each $$f_n$$ continuous, then $$f$$ is continuous. 1

• Follows from an $$\varepsilon/3$$ argument: \begin{align*} {\left\lvert {F(x) - F(y} \right\rvert} \leq {\left\lvert {F(x) - F_N(x)} \right\rvert} + {\left\lvert {F_N(x) - F_N(y)} \right\rvert} + {\left\lvert {F_N(y) - F(y)} \right\rvert} \leq {\varepsilon}\to 0 .\end{align*}

• The first and last $${\varepsilon}/3$$ come from uniform convergence of $$F_N\to F$$.
• The middle $${\varepsilon}/3$$ comes from continuity of each $$F_N$$.
• So just need to choose $$N$$ large enough and $$\delta$$ small enough to make all 3 $$\varepsilon$$ bounds hold.

If $$f_n \to f$$ uniformly, then $$\int f_n = \int f$$.

If $$f_n(x) \leq M_n$$ for a fixed $$x$$ where $$\sum M_n < \infty$$, then the series $$f(x) = \sum f_n(x)$$ converges pointwise. 2

If $$\sup_{x\in A} {\left\lvert {f_n(x)} \right\rvert} \leq M_n$$ for each $$n$$ where $$\sum M_n < \infty$$, then $$\sum_{n=1}^\infty f_n(x)$$ converges uniformly and absolutely on $$A$$. 3 Conversely, if $$\sum f_n$$ converges uniformly on $$A$$ then $$\sup_{x\in A} {\left\lvert {f_n(x)} \right\rvert} \to 0$$.

If $$E$$ is Lebesgue measurable, then $$E = H {\textstyle\coprod}N$$ where $$H \in F_\sigma$$ and $$N$$ is null.

For every $$\frac 1 n$$ there exists a closed set $$K_{n} \subset E$$ such that $$m(E\setminus K_{n}) \leq \frac 1 n$$. Take $$K = \cup K_{n}$$, wlog $$K_{n} \nearrow K$$ so $$m(K) = \lim m(K_{n}) = m(E)$$. Take $$N\coloneqq E\setminus K$$, then $$m(N) = 0$$.

Suppose $$E$$ is measurable; then for every $${\varepsilon}>0$$,

• There exists an open $$O\supset E$$ with $$m(O\setminus E) < {\varepsilon}$$
• There exists a closed $$F\subset E$$ with $$m(E\setminus F) < {\varepsilon}$$
• There exists a compact $$K\subset E$$ with $$m(E\setminus K) < {\varepsilon}$$.

• (1): Take $$\left\{{Q_{i}}\right\} \rightrightarrows E$$ and set $$O = \cup Q_{i}$$.
• (2): Since $$E^c$$ is measurable, produce $$O\supset E^c$$ with $$m(O\setminus E^c) < {\varepsilon}$$.
• Set $$F = O^c$$, so $$F$$ is closed.
• Then $$F\subset E$$ by taking complements of $$O\supset E^c$$
• $$E\setminus F = O\setminus E^c$$ and taking measures yields $$m(E\setminus F) < {\varepsilon}$$
• (3): Pick $$F\subset E$$ with $$m(E\setminus F) < {\varepsilon}/2$$.
• Set $$K_{n} = F\cap{\mathbb{D}}_{n}$$, a ball of radius $$n$$ about $$0$$.
• Then $$E\setminus K_{n} \searrow E\setminus F$$
• Since $$m(E) < \infty$$, there is an $$N$$ such that $$n\geq N \implies m(E\setminus K_{n}) < {\varepsilon}$$.

# Quintessential Qual Problems

• Prove the Lebesgue integral is translation/dilation invariant.
• Prove continuity in $$L_1$$: $${\left\lVert {\tau_hf - f} \right\rVert}\overset{h\to 0}\longrightarrow 0$$.
• Prove that $$E$$ is measurable $$\iff$$ $$E = F {\textstyle\coprod}Z$$ with $$F\in F_\sigma$$ and $$Z$$ null $$\iff$$ $$E = G\setminus Z$$ with $$G\in G_\delta$$ and $$Z$$ null.
• Show that $$m(E) = \sup_{K \subseteq E}m(K) \iff$$ there exists $$K = K({\varepsilon})$$ with $$m(K) \in [m(E) - {\varepsilon}, m(E)]$$.
• What’s most useful here is the proof technique, not so much the result itself.
• Apply Fubini and Tonelli to literally anything.
• Prove that $${\left\lVert {f} \right\rVert}_p\to {\left\lVert {f} \right\rVert}_\infty$$ over a finite measure space.
• Apply Cauchy-Schwarz to literally anything, in the form of $${\left\lVert {fg} \right\rVert}_1 \leq {\left\lVert {f} \right\rVert}_2 {\left\lVert {g} \right\rVert}_2$$.

Let $$E$$ be a measurable subset of $${\mathbf{R}}^n$$. Then

• For almost every $$x\in {\mathbf{R}}^{n_1}$$, the slice $$E_x \coloneqq\left\{{y \in {\mathbf{R}}^{n_2} \mathrel{\Big|}(x,y) \in E}\right\}$$ is measurable in $${\mathbf{R}}^{n_2}$$.

• The function

\begin{align*} F: {\mathbf{R}}^{n_1} &\to {\mathbf{R}}\\ x &\mapsto m(E_x) = \int_{{\mathbf{R}}^{n_2}} \chi_{E_x} ~dy \end{align*}

is measurable and \begin{align*} m(E) = \int_{{\mathbf{R}}^{n_1}} m(E_x) ~dx = \int_{{\mathbf{R}}^{n_1}} \int_{{\mathbf{R}}^{n_2}} \chi_{E_x} ~dy ~dx .\end{align*}

$$\implies$$:

• Let $$f$$ be measurable on $${\mathbf{R}}^n$$.
• Then the cylinders $$F(x, y) = f(x)$$ and $$G(x, y) = f(y)$$ are both measurable on $${\mathbf{R}}^{n+1}$$.
• Write $$\mathcal{A} = \left\{{G \leq F}\right\} \cap\left\{{G \geq 0}\right\}$$; both are measurable.

$$\impliedby$$:

• Let $$A$$ be measurable in $${\mathbf{R}}^{n+1}$$.
• Define $$A_x = \left\{{y\in {\mathbf{R}}\mathrel{\Big|}(x, y) \in \mathcal{A}}\right\}$$, then $$m(A_x) = f(x)$$.
• By the corollary, $$A_x$$ is measurable set, $$x \mapsto A_x$$ is a measurable function, and $$m(A) = \int f(x) ~dx$$.
• Then explicitly, $$f(x) = \chi_{A}$$, which makes $$f$$ a measurable function.
Footnotes
1.
Slogan: a uniform limit of continuous functions is continuous.
2.
Note that this is only pointwise convergence of $$f$$, whereas the full $$M{\hbox{-}}$$test gives uniform convergence.
3.
It suffices to show $${\left\lvert {f_n(x)} \right\rvert} \leq M_n$$ for some $$M_n$$ not depending on $$x$$.