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Let $(f_n)_{n\in\mathbb{N}}$ be a sequence of Riemann-integrable functions on $[a, b]$. Assume that $(f_n)_{n\in\mathbb{N}}$ converges pointwise to $f$ on $[a, b]$ and converges in the mean to $g$ on $[a, b]$, where $f$ and $g$ are both continuous. Show that $f = g$.

($(f_n)_{n\in\mathbb{N}}$ converges in the mean to g on $[a , b]$ means that $\lim_{n\to\infty} d(f_n, g) = 0$, where $d$ is the $L_2$ metric.)

I am studying real analysis now and want to prove this statement. But I don't know how to use condition "$(f_n)_{n\in\mathbb{N}}$ converges in the mean to g on $[a , b]$." How can I prove this?

Clement C.
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user509901
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2 Answers2

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Converges in mean implies that there is a subsequence $(f_{n_{k}})$ such that $f_{n_{k}}(x)\rightarrow g(x)$ pointwise almost everywhere, so $f=g$ a.e. and the continuity eliminates the a.e. condition.

user284331
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$d_{L_2}(f,g)\leq d_{L_2}(f,f_n)+d_{L_2}(f_n,g)$ by triangular inequality. If you can show that pointwise convergence implies convergence in $L_2$ metric, that is $d_{L_2}(f,f_n)\rightarrow 0$, then you will see $d_{L_2}(f,g)=0$, hence $f=g$ almost everywhere. Can you show

pointwise convergence implies convergence in $L_2$ metric

QED
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