Convergence and Continuity
\begin{align*} \limsup_n a_n = \lim_{n\to \infty} \sup_{j\geq n} a_j &= \inf_{n\geq 0} \sup_{j\geq n} a_j \\ \liminf_n a_n = \lim_{n\to \infty} \inf_{j\geq n} a_j &= \sup_{n\geq 0} \inf_{j\geq n} a_j .\end{align*}
A function \(f: {\mathbf{R}}\to {\mathbf{R}}\) is continuous on \(X\) iff for all \(x_0\in X\), \begin{align*} &\forall \varepsilon \quad \exists \delta(\varepsilon, x_0) \text{ such that }\quad \forall y, {\left\lvert {x_0  y} \right\rvert} < \delta &&\implies {\left\lvert {f(x_0)  f(y)} \right\rvert} < \varepsilon \\ \iff &\forall \varepsilon \quad \exists \delta(\varepsilon, x_0) \text{ such that }\quad \forall h, {\left\lvert {h} \right\rvert} < \delta &&\implies {\left\lvert {f(x_0)  f(x_0 \pm h)} \right\rvert} < \varepsilon .\end{align*}
\(f\) is uniformly continuous on \(X\) iff
\begin{align*} &\forall \varepsilon \quad \exists \delta(\varepsilon) \text{ such that }\quad \forall x, y, \in X \quad {\left\lvert {x  y} \right\rvert} < \delta &&\implies {\left\lvert {f(x)  f(y)} \right\rvert} < \varepsilon \\ \iff &\forall \varepsilon \quad \exists \delta(\varepsilon) \text{ such that} \quad \, \forall x, h, \quad {\left\lvert {h} \right\rvert} < \delta &&\implies {\left\lvert {f(x)  f(x \pm h)} \right\rvert} < \varepsilon .\end{align*} These follow from the substitutions \(x_0y = \mp h \implies y = x_0 \pm h\).
The main difference is that \(\delta\) may depend on \(x_0\) and \({\varepsilon}\) in continuity, but only depends on \({\varepsilon}\) in the uniform version. I.e. once \(\delta\) is fixed, for continuity one may only range over \(x\), but in uniform continuity one can range over all pairs \(x,y\).
Let \(X\) be a metric space and \(A\) a subset. Let \(A'\) denote the limit points of \(A\), and \(\overline{A} \coloneqq A\cup A'\) to be its closure.

A neighborhood of \(p\) is an open set \(U_p\) containing \(p\).

An \({\varepsilon}{\hbox{}}\)neighborhood of \(p\) is an open ball \(B_r(p) \coloneqq\left\{{q {~\mathrel{\Big\vert}~}d(p, q) < r}\right\}\) for some \(r>0\).

A point \(p\in X\) is an accumulation point or a limit point of \(A\) iff every punctured neighborhood \(U_p\setminus\left\{{p}\right\}\) contains a point \(q\in A\), so \(q\neq p\).

If \(p\in A\) and \(p\) is not a limit point of \(A\), then \(p\) is an isolated point of \(A\).

\(A\) is closed iff \(A' \subset A\), so \(A\) contains all of its limit points.

A point \(p\in A\) is interior iff there is a neighborhood \(U_p \subset A\) that is strictly contained in \(A\).

\(A\) is open iff every point of \(A\) is interior.

\(A\) is perfect iff \(A\) is closed and \(A\subset A'\), so every point of \(A\) is a limit point of \(A\).

\(A\) is bounded iff there is a real number \(M\) and a point \(q\in X\) such that \(d(p, q) < M\) for all \(p\in A\).

\(A\) is dense in \(X\) iff every point \(x\in X\) is either a point of \(A\), so \(x\in A\), or a limit point of \(A\), so \(x\in A'\). I.e., \(X\subset A\cup A'\).
 Alternatively, \(\overline{A} = X\), so the closure of \(A\) is \(X\).
A sequence of functions \(\left\{{ f_j }\right\}\) is said to converge pointwise to \(f\) if and only if \begin{align*} (\forall \varepsilon>0)(\forall x \in S)\left(\exists n_{0} = n_0(x, {\varepsilon}) \right)\left(\forall n>n_{0}\right)\left(\leftf_{n}(x)f(x)\right<\varepsilon\right) .\end{align*}
\begin{align*} (\forall \varepsilon>0)\left(\exists n_{0} = n_0({\varepsilon}) \right)(\forall x \in S)\left(\forall n>n_{0}\right)\left(\leftf_{n}(x)f(x)\right<\varepsilon\right) .\end{align*} Negated: ^{1} \begin{align*} (\exists \varepsilon>0)\left(\forall n_{0} = n_0 ({\varepsilon}) \right)(\exists x = x(n_0) \in S)\left(\exists n>n_{0}\right)\left(\leftf_{n}(x)f(x)\right \geq \varepsilon\right) .\end{align*}
Function Spaces
A metric space is complete if every Cauchy sequence converges.
If \(X\) is complete, then absolutely convergent implies convergent.
Recall that a set \(S\) in \(X\) is dense \(\iff\) every open \(U\subseteq X\) intersects \(S\). A set \(S\) is nowhere dense in \(X\) \(\iff\) the closure of \(S\) has empty interior \(\iff\) every subset (or interval) contains an open set (or a subinterval) that does not intersect \(S\). This just says \(S\) is not dense in any subset \(S' \subseteq X\), by negating what it means to be dense.
A set is meager if it is a countable union of nowhere dense sets.
A space \(X\) is a Baire space if and only if every countable intersections of open, dense sets is still dense.
Measure Theory
\begin{align*} \liminf_{n} E_{n} \coloneqq\bigcup_{N=1}^\infty \bigcap_{n=N}^\infty E_{n} &= \left\{{x {~\mathrel{\Big\vert}~}x\in E_{n} \text{ for all but finitely many } n}\right\} \\ \limsup_{n} E_{n} \coloneqq\bigcap_{N=1}^\infty \bigcup_{n=N}^{\infty} E_{n} &= \left\{{x {~\mathrel{\Big\vert}~}x\in E_{n} \text{ for infinitely many } n}\right\} .\end{align*}
How to derive these definitions: use that \(\inf\) corresponds to intersections/existence and \(\sup\) corresponds to unions/forall.

For \(\liminf E_n\):
 \(x\in \liminf E_n \iff\) there exists some \(N\) such that \(x\in \cap_{n\geq N} E_n\), i.e. \(x\in E_n\) for all \(n\geq N\). So \(x\) is in all but finitely many \(n\).
 How to remember: \(\liminf_{n} x_n = \sup_{n} \inf_{k\geq n} x_n\) for sequences, where sups look like unions and infs look like intersections.
 Alternatively: there exists an \(n\) (union) such that for all \(k\geq n\) (intersection)…

For \(\limsup E_n\):
 \(x\in \limsup E_n \iff\) for every \(N\), there exists some \(n\geq N\) such that \(x\in E_n\). So \(x\) is an infinitely many \(E_n\).
 How to remember: \(\limsup_{n} x_n = \inf{n} \sup{k\geq n} x_n\) for sequences, where sups look like unions and infs look like intersections.
 Alternatively: for all \(n\) (intersection) there exists a \(k\geq n\) (union)…
It’s also useful to note that \(\liminf E_n \subseteq \limsup E_n\), since \(\liminf E_n\) are elements that are eventually in all sets, and \(\limsup E_n\) are elements in infinitely many sets.
Why these are useful: for finite measure spaces, \begin{align*} \mu\qty{\liminf_n E_n }\leq \liminf_n \mu(E_n) \leq \lim_n \mu(E_n) \leq \limsup_n \mu(E_n) \leq \mu\qty{\limsup_n E_n} .\end{align*} If the \(\limsup\) and \(\liminf\) sets are equal, then one can define the set \(\lim_n E_n \coloneqq\cup_n E_n\) if \(E_n \nearrow E\) or \(\lim_n E_n \coloneqq\cap_n E_n\) if \(E_n\searrow E\) in which case continuity of measure states \begin{align*} \mu\qty{\lim_n E_n} = \lim_n \mu(E_n) .\end{align*}
An \(F_\sigma\) set is a union of closed sets, and a \(G_\delta\) set is an intersection of opens. ^{2}
The outer measure of a set is given by \begin{align*} m_*(E) \coloneqq\inf_{\substack{\left\{{Q_{i}}\right\} \rightrightarrows E \\ \text{closed cubes}}} \sum {\left\lvert {Q_{i}} \right\rvert} ,\end{align*} where \({\left\lvert {Q_i} \right\rvert}\) is the standard Euclidean volume of a cube in \({\mathbf{R}}^n\).
A subset \(E\subseteq {\mathbf{R}}^n\) is Lebesgue measurable iff for every \({\varepsilon}> 0\) there exists an open set \(O \supseteq E\) such that \(m_*(O\setminus E) < {\varepsilon}\). In this case, we define \(m(E) \coloneqq m_*(E)\).
\(f\in L^+\) iff \(f\) is measurable and nonnegative.
Integrals and \(L^p\) Spaces
A measurable function is integrable iff \({\left\lVert {f} \right\rVert}_1 < \infty\).
\begin{align*} {\left\lVert {f} \right\rVert}_\infty &\coloneqq\inf_{\alpha \geq 0} \left\{{\alpha {~\mathrel{\Big\vert}~}\mu\qty{\left\{{{\left\lvert {f} \right\rvert} \geq \alpha}\right\}} = 0}\right\} .\end{align*} In words, this is the smallest upper bound that holds almost everywhere, so \({\left\lvert {f(x)} \right\rvert} \leq {\left\lVert {f} \right\rVert}_\infty\) holds for almost every \(x\). A function \(f:X \to {\mathbf{C}}\) is essentially bounded iff there exists a real number \(c\) such that \(\mu(\left\{{{\left\lvert {f} \right\rvert} > x}\right\}) = 0\), i.e. \({\left\lVert {f} \right\rVert}_\infty < \infty\).
\begin{align*} L^\infty(X) \coloneqq\left\{{f: X\to {\mathbf{C}}{~\mathrel{\Big\vert}~}f \text{ is essentially bounded }}\right\} \coloneqq\left\{{f: X\to {\mathbf{C}}{~\mathrel{\Big\vert}~}{\left\lVert {f} \right\rVert}_{\infty }< \infty}\right\} .\end{align*}
\begin{align*}f * g(x)=\int f(xy) g(y) d y .\end{align*}
\begin{align*} \widehat{f}(\xi) = \int f(x) ~e^{2\pi i x \cdot \xi} ~dx .\end{align*}
\begin{align*} \phi_{t}(x) = t^{n} \phi\left(t^{1} x\right) .\end{align*}
For \(\phi\in L^1\), the dilations satisfy \(\int \phi_{t} = \int \phi\), and if \(\int \phi = 1\) then \(\phi\) is an approximate identity.
Some properties that approximate identities enjoy:
 \(\int \phi_t = {\left\lVert {\phi_t} \right\rVert}_1 = 1\) for every \(t\).
 \(\sup_t {\left\lVert {\phi_t} \right\rVert}_1 < \infty\)
 For every \(h>0\), \(\lim_{t\to\infty} \int_{{\left\lvert {x} \right\rvert} \geq h} {\left\lvert {\phi_t(x)} \right\rvert}\,dx= 0\).
Functional Analysis
For \(X\) a normed vector space and \(L \in X {}^{ \vee }\), the dual norm or operator norm is defined by \begin{align*} {\left\lVert {L} \right\rVert}_{X {}^{ \vee }} \coloneqq\sup_{ \substack{x\in X \\ {\left\lVert {x} \right\rVert} = 1} } {\left\lvert {L(x)} \right\rvert} = \sup_{ \substack{x\in X \\ {\left\lVert {x} \right\rVert} \leq 1} } {\left\lvert {L(x)} \right\rvert} .\end{align*}
A countable collection of elements \(\left\{{ u_i }\right\}\) is orthonormal if and only if
 \({\left\langle {u_i},~{u_j} \right\rangle} = 0\) for all \(j \neq k\) and
 \({\left\lVert {u_j} \right\rVert}^2 \coloneqq{\left\langle {u_j},~{u_j} \right\rangle} = 1\) for all \(j\).
A set \(\left\{{u_{n}}\right\}\) is a basis for a Hilbert space \({\mathcal{H}}\) iff it is dense in \({\mathcal{H}}\).
A collection of vectors \(\left\{{u_{n}}\right\}\subset H\) is complete iff \({\left\langle {x},~{u_{n}} \right\rangle} = 0\) for all \(n \iff x = 0\) in \(H\).
The dual of a Hilbert space \(H\) is defined as \begin{align*} H {}^{ \vee }\coloneqq\left\{{L: H\to {\mathbf{C}}{~\mathrel{\Big\vert}~}L \text{ is continuous }}\right\} .\end{align*}
A map \(L: X \to {\mathbf{C}}\) is a linear functional iff \begin{align*} L(\alpha\mathbf{x} + \mathbf{y}) = \alpha L(\mathbf{x}) + L(\mathbf{y}). .\end{align*}
A space is a Banach space if and only if it is a complete normed vector space.
A Hilbert space is an inner product space which is a Banach space under the induced norm.