Convergence in measure implies $L^1$ norm-(Vitali Convergence Theorem)

Question

Let $(X, \nu)$ be a finite measure space and let $f_n$ be a sequence of integrable functions on $X$ converging in measure to a function $f$. Suppose that $\int |f_n|^p d\nu\leq A$ for some constant $A$ and some $p > 1$. Show that then $f_n \to f$ in the $L^1$ norm.

Answer 1

It is something related to uniform integrability. The following theorems and the notion of uniform integrability are due to Vitali, which play an important role in probability theory. For your problem, solution follows directly from the following theorems.

Definition: Let $(X,\mathcal{M},\mu)$ be a measure space with $\mu(X)<\infty$. Let $\mathcal{C}$ be a family of measurable functions defined on $X$. We say that $\mathcal{C}$ is uniformly integrable if for each $\varepsilon>0$, there exists $c>0$ such that $$ \int_{\{x\mid|f(x)|\geq c\}}|f|d\mu<\varepsilon $$ for any $f\in\mathcal{C}$. We remark that if $\mathcal{C}$ is a finite family of integrable functions, $\mathcal{C}$ is automatically uniformly integrable.

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Proposition 1: Let $(X,\mathcal{M},\mu)$ be a measure space with $\mu(X)<\infty$. Let $\mathcal{C}$ be a family of measurable functions defined on $X$. The following are equivalent:

(a) $\mathcal{C}$ is uniformly integrable.

(b) $\sup_{f\in\mathcal{C}}\|f\|_{1}<\infty$ and for each $\varepsilon>0$, there exists $\delta>0$ such that $\int_{A}|f|d\mu<\infty$ whenever $f\in\mathcal{C}$ and $\mu(A)<\delta$.

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Theorem 2: Let $(X,\mathcal{M},\mu)$ be a measure space with $\mu(X)<\infty$. Let $\mathcal{C}$ be a family of measurable functions. Let $p\in(1,\infty)$. If $\sup_{f\in\mathcal{C}}\|f\|_{p}<\infty$, then $\mathcal{C}$ is uniformly integrable.

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Theorem 3 (Vitali): Let $(X,\mathcal{M},\mu)$ be a measure space with $\mu(X)<\infty$. Let $(f_{n})$ be a sequence of integrable functions and let $f$ be an integrable function. Then the followings are equivalent:

(a) $\|f_{n}-f\|_{1}\rightarrow0$.

(b) $f_{n}\rightarrow f$ in measure $\mu$ and the family $\{f_{n}\mid n\in\mathbb{N}\}$ is uniformly integrable.

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Proof of Proposition 1:

$(a)\Rightarrow(b):$ Suppose that $\mathcal{C}$ is uniformly integrable. Take $\varepsilon=1$, then there exists $c>0$ such that $\int_{[|f|\geq c]}|f|<1$ for all $f\in\mathcal{C}$. Now \begin{eqnarray*} \int|f| & = & \int_{[|f|\geq c]}|f|+\int_{[|f|<c]}|f|\\ & \leq & 1+c\mu(X). \end{eqnarray*} Hence $\sup_{f\in\mathcal{C}}\|f\|_{1}<\infty$. Note for any integrable function $g$ and $c>0$, we have Markov's inequality: $c\mu\left(\left[|g|\geq c\right]\right)\leq\int_{[|g|\geq c]}|g|\leq\int|g|$. Let $M=\sup_{f\in\mathcal{C}}\|f\|_{1}<\infty$. Let $\varepsilon>0$ be arbitrary. By uniform integrability of $\mathcal{C}$, there exists $c>0$ such that $\int_{[|f|\geq c]}|f|<\frac{\varepsilon}{2}$ whenever $f\in\mathcal{C}$. Let $\delta=\frac{\varepsilon}{2c}$. Let $f\in\mathcal{C}$ and $A\in\mathcal{M}$ be arbitrary with $\mu(A)<\delta$. Then \begin{eqnarray*} & & \int_{A}|f|\\ & = & \int_{A\cap[|f|\geq c]}|f|+\int_{A\cap[|f|<c]}|f|\\ & < & \frac{\varepsilon}{2}+c\mu(A)\\ & < & \frac{\varepsilon}{2}+c\delta\\ & = & \varepsilon. \end{eqnarray*}

$(b)\Rightarrow(a):$ Let $M=\sup_{f\in\mathcal{C}}\|f\|_1<\infty$. Let $\varepsilon>0$ be given. Choose $\delta>0$ such that $\int_{A}|f|<\varepsilon$ whenever $f\in\mathcal{C}$ and $\mu(A)<\delta$. Let $c=\frac{M+1}{\delta}>0$. For any $f\in\mathcal{C}$, by Markov's inequality, we have $\mu\left([|f|\geq c]\right)\leq\frac{1}{c}M<\delta$. It follows that $\int_{[|f|\geq c]}|f|<\varepsilon$.

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Proof of Theorem 2:

Let $q\in(1,\infty)$ be such that $\frac{1}{p}+\frac{1}{q}=1$. Let $M=\sup_{f\in\mathcal{C}}\|f\|_{p}<\infty$. By Holder inequality, for any $f\in\mathcal{C}$, we have \begin{eqnarray*} \int|f| & = & \int|f|\cdot1\\ & \leq & \|f\|_{p}\|1\|_{q}\\ & \leq & M\mu(X)^{\frac{1}{q}}. \end{eqnarray*} Therefore $\sup_{f\in\mathcal{C}}\|f\|_{1}<\infty$. Let $\varepsilon>0$ be given. Choose $\delta>0$ such that $M\delta^{\frac{1}{q}}<\varepsilon.$ For any $f\in\mathcal{C}$ and $A\in\mathcal{M}$ with $\mu(A)<\delta$, by Holder inequality, we have: \begin{eqnarray*} & & \int_{A}|f|\\ & \leq & \|f\|_{p}\|1_{A}\|_{q}\\ & \leq & M\mu(A)^{^{\frac{1}{q}}}\\ & \leq & M\delta^{\frac{1}{q}}\\ & < & \varepsilon. \end{eqnarray*} By Proposition 1, it follows that $\mathcal{C}$ is uniformly integrable.

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Proof of Theorem 3: $(a)\Rightarrow(b)$: Suppose that $\|f_{n}-f\|_{1}\rightarrow0$. Let $c>0$. By Markov inequality, we have $\mu\left([|f_{n}-f|\geq c]\right)\leq\frac{1}{c}\|f_{n}-f\|_{1}\rightarrow0$ as $n\rightarrow\infty$. Therefore $f_{n}\rightarrow f$ in measure $\mu$. Clearly $\sup_{n}\|f_n-f\|_{1}<\infty$. Let $\varepsilon>0$ be given. For each $n$, choose $\delta_{n}>0$ such that $\int_{A}|f_{n}-f|<\varepsilon$ whenever $\mu(A)<\delta_{n}$. Choose $N$ such that $\|f_{n}-f\|_{1}<\varepsilon$ whenever $n>N$. Define $\delta=\min(\delta_{1},\ldots,\delta_{N})>0$. Let $n\in\mathbb{N}$ and $A\in\mathcal{M}$ with $\mu(A)<\delta$. If $n<N$, we have $\int_{A}|f_{n}-f|<\varepsilon$ because $\mu(A)<\delta\leq\delta_{n}$. If $n\geq N$, we have $\int_{A}|f_{n}-f|\leq\|f_{n}-f\|_{1}<\varepsilon$. This shows that $\{f_{n}-f\mid n\in\mathbb{N}\}$ is uniformly integrable (by Proposition 1). Lastly, since $f$ is integrable and $f_{n}=(f_{n}-f)+f$, it is easy to see that $\{f_{n}\mid n\in\mathbb{N}\}$ is uniformly integrable.

$(b)\Rightarrow(a):$ Suppose that $f_{n}\rightarrow f$ in measure and that $\{f_{n}\mid n\in\mathbb{N}\}$ is uniformly integrable. Let $g_{n}=f_{n}-f$. Clearly $\{g_{n}\mid n\in\mathbb{N}\}$ is uniformly integrable. Let $\varepsilon>0$ be given, then there exists $\delta>0$ such that $\int_{A}|g_{n}|<\varepsilon/2$ whenver $\mu(A)<\delta$ and $n\in\mathbb{N}$. Choose $c>0$ such that $c\mu(X)<\varepsilon/2$. Since $f_{n}\rightarrow f$ in measure, there exists $N$ such that $\mu\left([|g_{n}|\geq c]\right)<\delta$ whenever $n\geq N$. For any $n\geq N$, we have \begin{eqnarray*} & & \int|g_{n}|\\ & = & \int_{[|g_{n}|\geq c]}|g_{n}|+\int_{[|g_{n}|<c]}|g_{n}|\\ & < & \frac{\varepsilon}{2}+c\mu(X)\\ & < & \varepsilon. \end{eqnarray*} In the above, $\int_{[|g_{n}|\geq c]}|g_{n}|<\varepsilon/2$ because $\mu\left([|g_{n}|\geq c]\right)<\delta$.