
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 CauchySchwarz 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.