Advice and Essentials

  • 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 lim sup 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

proposition (Convergent Sums Have Small Tails):

\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*}

theorem (Uniform Limit Theorem):

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

proof:

    
    
      
        
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      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*}

    
    
  
    
      
      
        
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          The first and last {\varepsilon}/3 come from uniform convergence of F_N\to F.
        
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          The middle {\varepsilon}/3 comes from continuity of each F_N.
        
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      So just need to choose N large enough and \delta small enough to make all 3 \varepsilon bounds hold.
    
    
  
        
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proposition (Uniform Limits Commute with Integrals):
    

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

  
proposition (Weak M\dashTest):
    

      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
    
  
    

  
proposition (The Weierstrass M\dashTest):
    

      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.
    

  
proposition (Borel Characterization of Measurable Sets):
    

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

  
proof (of Borel characterization):
    

      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.
    

  
theorem (Measurable sets can be approximated by open/closed/compact sets.):
    

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

  
    
    
      
        
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          There exists an open O\supset E with m(O\setminus E) < {\varepsilon}
        
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          There exists a closed F\subset E with m(E\setminus F) < {\varepsilon}
        
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          There exists a compact K\subset E with m(E\setminus K) < {\varepsilon}.
        
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proof (that measurable sets can be approximated):

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

    
    
      
        
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          Prove the Lebesgue integral is translation/dilation invariant.
        
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          Prove continuity in L_1: {\left\lVert {\tau_hf - f} \right\rVert}\overset{h\to 0}\longrightarrow 0.
        
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          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.
        
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          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)].
    
      
      
        
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          What’s most useful here is the proof technique, not so much the result itself.
        
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          Apply Fubini and Tonelli to literally anything.
        
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          Prove that {\left\lVert {f} \right\rVert}_p\to {\left\lVert {f} \right\rVert}_\infty over a finite measure space.
        
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          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.
        
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proposition (Measurable Slices):
    

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

  
    
    
      
        
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      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}.
    
    
  
        
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      The function
    
    
  
        
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\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*}

    

  
proof (of measurable slices):

    

      \implies:
    

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

  
    
    
      
        
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          Let A be measurable in {\mathbf{R}}^{n+1}.
        
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          Define A_x = \left\{{y\in {\mathbf{R}}\mathrel{\Big|}(x, y) \in \mathcal{A}}\right\}, then m(A_x) = f(x).
        
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          By the corollary, A_x is measurable set, x \mapsto A_x is a measurable function, and m(A) = \int f(x) ~dx.
        
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          Then explicitly, f(x) = \chi_{A}, which makes f a measurable function.
        
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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.