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This is not true for infinite measures (Pointwise convergence, but not in measure). Is it true for a finite measure? Namely, let a finite (probability) measure $\mu(\cdot)$. Does a point-wise convergence of $\mu-$measurable functions, $\{f_n(x)\}$, imply convergence in measure $\mu(\cdot)$?

Equivalently, does $$ \forall x~\lim_{n\to \infty} f_n(x) \to f(x)\,, $$ imply $$ \forall \varepsilon ~\lim_{n\to \infty} \mu\big{(}\left\{x~\big{|}~|(f_n(x) - f(x)|>\varepsilon\right\}\big{)} \to 0\,~? $$

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3 Answers 3

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Yeah this is true. You can use Egoroff, but a direct proof using only DCT is not difficult either.

Let $\epsilon>0$. Define $A_n:=\{x: |f(x)-f_n(x)|>\epsilon\}$. Since $f_n \to f$ pointwise, it follows that $1_{A_n} \to 0$ pointwise. Then notice that all of the functions $1_{A_n}$ are bounded above by the constant function $1$, which is in $ L^1$ (since $\mu$ is finite). By DCT, it follows that $\mu(A_n) = \int 1_{A_n} d\mu \to 0$. But $\mu(A_n) \to 0$ precisely means that $f_n \to f$ in measure.

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If $f_n \overset{a.e.}{\longrightarrow} f$ and the space has finite measure, by Egoroff's theorem $f_n$ converges to $f$ almost uniformly, and therefore converges in measure to $f$.

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By the reversed Fatou lemma, we have

$\lim \mu( |f_n-f|\geq \epsilon) = \lim \int 1_{\{ |f_n-f|\geq \epsilon \}}d\mu \leq \int \limsup 1_{\{|f_n-f|\geq \epsilon \}}d\mu = 0.$

$1_{\{\cdot\}}$ is the indicator function (or characteristic function for analysis people). Since indicator is bounded, it is integrable as the measure is finite. The use of Fatou lemma is then justified.

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