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Lebesgue integration exercises

 
  • Jan 06, 2023
    1. Give an example of a sequence $\left(f_n\right)$ in $β„’^1(ℝ)$ such that $\lim_{nβ†’βˆž}\left\|f_n\right\|_1=0$ and $\limsup_{nβ†’βˆž}f_n(x)=∞$ for each $x∈(0,1)$.
      $f_n(x)=rΟ‡_{((k-1)2^{-r},k2^{-r})},n=2^r+k,k∈\{1,2,…,2^r\}$
    2. βˆ„ a sequence $\left(g_n\right)_{nβ©Ύ1}$ in $β„’^1(ℝ)$ such that $\lim_{nβ†’βˆž}\left\|g_n\right\|_1=0$ and $\liminf_{nβ†’ ∞}\left|g_n(x)\right|>0$ for each $x∈(0,1)$
      Proof: By Fatou's lemma$$\int_ℝ\liminf_{nβ†’βˆž}{|g_n(x)|}β©½\liminf_{nβ†’βˆž}\int_ℝ{|g_n(x)|}=0$$so $\liminf_{nβ†’ ∞}\left|g_n(x)\right|=0$ a.e.
    1. State the Monotone and the Dominated Convergence Theorems.
      Theorem 4.2. [Monotone Convergence Theorem] If $(f_n)$ is an increasing sequence of non-negative measurable functions and $f=\lim_{nβ†’βˆž}f_n$, then $∫f=\lim_{nβ†’βˆž}∫f_n$
      Theorem 6.3. [Dominated Convergence Theorem] Let $(f_n)$ be a sequence of measurable functions such that:
      1. $(f_n(x))$ converges a.e. to a limit $f(x)$,
      2. there is an integrable function $g$ such that, for each $n$, ${|f_n(x)|}≀g(x)$ a.e.
      Then $f$ is integrable, and $∫f=\lim_{nβ†’βˆž}∫f_n$
    2. Let $\left(f_n\right)$ be a sequence of real-valued Lebesgue integrable functions on ℝ. Prove that if the series$$\sum_{n=1}^∞\int_ℝ\left|f_n\right|$$is convergent to $Kβˆˆβ„$ then the series $\sum_{n=1}^∞\left|f_n\right|$ converges almost everywhere to an integrable function.
      For every $x$, the sequence $g_m(x)≔\sum_{n=1}^m\left|f_n(x)\right|,m=1,2,…$ is increasing\[\int_ℝg_m=\sum_{n=1}^m\int_ℝ\left|f_n\right|β©½\sum_{n=1}^∞\int_ℝ\left|f_n\right|=K\]So $g_m(x)<∞$ a.e. $x$ [Bounded monotone sequence converges] So $g_m(x)$ converges a.e. $x$\[\int_ℝ\lim_{mβ†’βˆž}g_m\xlongequal{\text{MCT}}\lim_{mβ†’βˆž}\int_ℝg_m\xlongequal{\text{interchange ∫ with finite sum}}\lim_{mβ†’βˆž}\sum_{n=1}^m\int_ℝ\left|f_n\right|=K\]
      Deduce, or prove otherwise, that $\sum_{n=1}^∞f_n$ converges almost everywhere to an integrable function $h$ and that $$ \int_ℝ h=\sum_{n=1}^∞\int_ℝ f_n $$
      We just proved $\sum_{n=1}^∞\left|f_n\right|$ converges a.e. to a function $g$, so [Absolute convergence implies convergence] $\sum_{n=1}^∞f_n$ converges a.e. to a function $h$. $$\left|h_m\right|≀g_m≀g⟹\int_ℝh\xlongequal{\text{DCT}}\lim_{mβ†’βˆž}\int_ℝh_m\xlongequal{\text{interchange ∫ with finite sum}}\lim_{mβ†’βˆž}\sum_{n=1}^m\int_ℝf_m=\sum_{n=1}^∞\int_ℝ f_n$$
    3. Prove that if $α$ is a real number then $$ \int_0^∞\frac{\sin α x}{\mathrm{e}^x-1} \mathrm{~d} x=\sum_{n=1}^∞\frac{α}{α^2+n^2} . $$
      For each positive integer $n$ the function $Ο•_n(x)=Ο‡_{(0, ∞)}(x) x \mathrm{e}^{-n x}$ and $f_n(x):=Ο‡_{(0, ∞)}(x) \mathrm{e}^{-n x} \sin Ξ± x$ are integrable.
      For $xβ‰₯0$, ${|\sin Ξ±x|}≀{|Ξ±|}x$. So $\left|f_n(x)\right|≀{|Ξ±|}Ο•_n(x)$.$$\sum_{n=1}^∞\int_ℝϕ_n(x)=\sum_{n=1}^∞\frac1{n^2}<∞$$So the condition in (b) is satisfied. $$ \int_0^∞\frac{\sinΞ± x}{e^x-1}=\int_ℝ\sum_{n=1}^∞f_n\overset{\text{(b)}}=\sum_{n=1}^∞\int_ℝf_n=\sum_{n=1}^∞\fracΞ±{Ξ±^2+n^2} $$
    1. Give careful statements of the theorems of Fubini and Tonelli.
      Theorem 8.2. [Fubini’s Theorem] Let $f:ℝ^2→ℝ$ be integrable. Then, for almost all $y$, the function $x↦f(x,y)$ is integrable. Moreover, defining (for almost all $y$) by $F(y)=∫f(x,y)dx$, then $F$ is integrable, and$$\int_{ℝ^{2}} f(x, y) d(x, y)=\int_ℝ\left(\int_ℝ f(x, y) d x\right) d y$$
      Theorem 8.3. [Tonelli’s Theorem] Let $f:ℝ^2→ℝ$ be a measurable function, and suppose that either of the following repeated integrals is finite:$$\int_ℝ\left(\int_ℝ|f(x, y)| d x\right) d y, \quad \int_ℝ\left(\int_ℝ|f(x, y)| d y\right) d x$$Then $f$ is integrable.
    2. By applying these theorems to the function $f$ defined on $[0,n]Γ—[0,∞)$ by $$ f(x, t)=\sin x \mathrm{e}^{-t x} $$ show that $$ \lim_{nβ†’βˆž}\int_0^n \frac{\sin x}{x}\mathrm{~d}x=\fracΟ€2 $$
      \begin{align*} I_n&=\int_0^n\int_0^∞\sin x\mathrm{e}^{-t x}\mathrm{~d}t\mathrm{~d}x\\ &=\int_0^n\frac{\sin x}x\mathrm{~d}x \end{align*} ${|f(x,t)|}≀ x \mathrm{e}^{-t x}$ and $∫_0^n∫_0^ ∞x\mathrm{e}^{-t x}\mathrm{~d}t\mathrm{~d}x=∫_0^n1\mathrm{~d}x=n$, by Tonelli $f(x,t)$ is integrable on $[0,n]Γ—[0,∞)$. By Fubini, \begin{align*} I_n&=\int_0^∞\int_0^n\sin x\mathrm{e}^{-t x}\mathrm{~d}x\mathrm{~d}t\\ &=\int_0^∞\frac1{t^2+1}\mathrm{~d}t-\int_0^∞\frac{e^{-tn}\left(t\sin n+\cos n\right)}{t^2+1}\mathrm{~d}t \end{align*} Taking the limit $nβ†’βˆž$ the second integral$β†’0$ [Because $\left|\frac{e^{-tn}\left(t\sin n+\cos n\right)}{t^2+1}\right|≀(t+1)e^{-tn}$ and $∫_0^∞(t+1)e^{-tn}\mathrm dt=n^{-2}+n^{-1}β†’0$] \[\lim_{nβ†’βˆž}I_n=\int_0^∞\frac1{t^2+1}\mathrm{~d}t=\fracΟ€2\]
    3. Is $\frac{\sin x}{x}$ Lebesgue integrable on $(0, ∞)$ ?
      Ex5.3 5. Consider $f(x)=(\sin x) / x$ over $(0, ∞)$. Now $$ \int_{r \pi}^{(r+1) \pi}\left|\frac{\sin x}{x}\right| d x \geq \int_{r \pi}^{(r+1) \pi} \frac{|\sin x|}{(r+1) \pi} d x=\frac{2}{(r+1) \pi} . $$ Hence, $$ \lim_{nβ†’βˆž}\int_0^{n \pi}{|f(x)|} d x \geq \lim_{nβ†’βˆž}\sum_{r=0}^{n-1} \frac{2}{(r+1) \pi}=∞ . $$ So $|f|$ is not integrable, and hence $f$ is not integrable, over $(0, ∞)$.