\(L^1\)

For \(f\in L^+\), \begin{align*} \int f = 0 \quad\iff\quad f \equiv 0 \text{ almost everywhere} .\end{align*} 


    
    
      
        
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          Obvious for simple functions:
    
      
      
        
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          If \(f(x) = \sum_{j=1}^n c_j \chi_{E_j}\), then \(\int f = 0\) iff for each \(j\), either \(c_j=0\) or \(m(E_j) = 0\).
        
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          Since nonzero \(c_j\) correspond to sets where \(f\neq 0\), this says \(m\qty{\left\{{f\neq 0}\right\}} = 0\).
        
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          \(\impliedby\):
    
      
      
        
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          If \(f= 0\) almost everywhere and \(\phi \nearrow f\), then \(\phi = 0\) almost everywhere since \(\phi(x) \leq f(x)\) -Then 

\begin{align*}  
\int f = \sup_{\phi \leq f} \int \phi = \sup_{\phi \leq f} 0 = 0
.\end{align*}

        
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          \(\implies\):
    
      
      
        
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          Instead show negating “\(f=0\) almost everywhere” implies \(\int f \neq 0\).
        
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          Write \(\left\{{f\neq 0}\right\} = \cup_{n\in {\mathbb{N}}} S_n\) where \(S_n \coloneqq\left\{{x{~\mathrel{\Big\vert}~}f(x) \geq {1\over n}}\right\}\).
        
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          Since “not \(f=0\) almost everywhere”, there exists an \(n\) such that \(m(S_n) > 0\).
        
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          Then 

\begin{align*}  
0 < {1\over n} \chi_{E_n} \leq f \implies 
0 < \int {1\over n} \chi_{E_n} \leq \int f
.\end{align*}

        
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      The Lebesgue integral is translation invariant, i.e. 

\begin{align*}
\int f(x) ~dx = \int f(x + h) ~dx &&\text{ for any } h
.\end{align*}

    

  

    
    
      
        
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          Let \(E\subseteq X\); for characteristic functions, 

\begin{align*}
\int_X \chi_E(x+h) 
= \int_{X} \chi_{E+h}(x) = m(E+h) = m(E) = \int_X \chi_E(x)
\end{align*}
 by translation invariance of measure.
        
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          So this also holds for simple functions by linearity.
        
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          For \(f\in L^+\), choose \(\phi_n \nearrow f\) so \(\int \phi_n \to \int f\).
        
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          Similarly, \(\tau_h \phi_n \nearrow \tau_h f\) so \(\int \tau_h f \to \int f\)
        
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          Finally \(\left\{{\int \tau_h \phi}\right\} = \left\{{\int \phi}\right\}\) by step 1, and the suprema are equal by uniqueness of limits.
        
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\begin{align*}
X \subseteq A {\textstyle\coprod}B \implies \int_X f &\leq \int_A f + \int_B f\\
X = A {\textstyle\coprod}B \implies \int_X f &= \int_A f + \int_B f
.\end{align*}

    

  

    

      

\begin{align*}
f\in L^1 \text{ uniformly continuous }\implies f(x) \overset{{\left\lvert {x} \right\rvert}\to \infty }\longrightarrow 0
.\end{align*}

    

  

    

      This doesn’t hold for general \(L^1\) functions, take any train of triangles with height 1 and summable areas.
    

  

    

      

\begin{align*}
f\in L^1({\mathbf{R}}^d) \implies {\left\lVert {f} \right\rVert}_{L^1(B_r(0)^c)} \overset{r\to\infty}\longrightarrow 0
.\end{align*}
 In particular, if \(f\in L^1\), then for every \(\varepsilon\) there exists a radius \(r\) such that 

\begin{align*}
\int_{B_r(0)^c} {\left\lvert {f} \right\rvert} < {\varepsilon}
.\end{align*}

    

  

    
    
      
        
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          Approximate with compactly supported functions.
        
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          Take \(g\overset{L_1}\to f\) with \(g\in C_c\)
        
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          Then choose \(N\) large enough so that \(g=0\) on \(E\coloneqq B_N(0)\)
        
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          Then 

\begin{align*} \int_E {\left\lvert {f} \right\rvert} \leq \int_E{\left\lvert {f-g} \right\rvert} + \int_E {\left\lvert {g} \right\rvert}.\end{align*}

        
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\begin{align*}
f\in L^1 \implies \int_E f \overset{m(E) \to 0}\longrightarrow 0
.\end{align*}

    

  

    

      Approximate with compactly supported functions. Take \(g\overset{L_1}\to f\), then \(g \leq M\) so \(\int_E{f} \leq \int_E{f-g} + \int_E g \to 0 + M \cdot m(E) \to 0\).
    

  

    

      

\begin{align*}
f\in L^1 \implies 
m(\left\{{f(x) = \infty}\right\}) = 0
.\end{align*}

    

  

    

      Alternate idea: Split up domain Let \(A = \left\{{f(x) = \infty}\right\}\), then \(\infty > \int f = \int_A f + \int_{A^c} f = \infty \cdot m(A) + \int_{A^c} f \implies m(X) =0\).
    

  

    

      

\begin{align*} 
f\in L^1 \implies {\left\lVert {\tau_h f - f} \right\rVert}_1 \overset{h\to 0}\to 0
\end{align*}
 Interesting related facts:
    

  
    
    
      
        
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          \(\left\{{\tau_h {~\mathrel{\Big\vert}~}h\in {\mathbf{R}}}\right\}\) is equicontinuous.
        
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          By Ascoli, this locally uniformly converges (so uniformly on all compact subsets)
        
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      Prove for uniformly continuous function. Then clear for \(C_c^0\) by \({\varepsilon}/3\), using uniform continuity to bound \({\left\lVert {\tau_h g- g} \right\rVert}\): 

\begin{align*}
{\left\lVert {\tau_hf - f} \right\rVert} 
&= {\left\lVert {\tau_hf - \tau_hg + \tau_h g + g - g - f} \right\rVert}
.\end{align*}

    

  
    

      
    

      See https://math.stackexchange.com/questions/458230/continuity-of-l1-functions-with-respect-to-translation
    

  
    

  

    

      

\begin{align*}
A_{h}(f)(x):=\frac{1}{2 h} \int_{x-h}^{x+h} f(y) d y
\implies {\left\lVert {A_h(f) - f} \right\rVert} \overset{h\to 0}\to 0
.\end{align*}

    

  

    

      Fubini-Tonelli, and sketch region to change integration bounds, and continuity in \(L^1\).
    

  

    

      

\begin{align*}
{\left\langle {f},~{\phi} \right\rangle} \coloneqq\int f \cdot \phi = 0 \quad \forall \phi\in C_c^0 \implies f \underset{{ \text{a.e.} }}{=}0
.\end{align*}

    

  

    

      Let \(A\) be an interval, choose \(\phi_k \to \chi_A\), then \(\int f \chi_A = 0\) for all intervals. So this holds for any Borel set \(A\). Then just take \(A_1 = \left\{{f > 0}\right\}\) and \(A_2 = \left\{{f < 0}\right\}\), then \(\int_{\mathbf{R}}f = \int_{A_1} f + \int_{A_2}f = 0\).