poinwise limit and limit in convergence is same if they are continuous

Question

let ${f_k}$ be a sequence of integrable functions on $[a,b]$ that converges in pointwise to $f$ and converges in mean to $g$. If $f$ and $g$ are continuous, then f = g. The 'convergence in mean' means, $\lim_{k\to \infty}\int_{a}^{b} [f_k(x) - g(x)]^2 dx = 0$ .

I tried to show that, if there exists some point $x_0$ s.t $f(x_0) \neq g(x_0)$, then on some neighborhood of $x_0$, $f_k-g$ is being uniformly nonzero, so that the value of integration can't be 0, so that $f_k$ can't converge in mean to $g$.

However, I couldn't prove this way, since and $f_k$ is integrable, but not provided that $f_k$ is continuous.

Am I trying in a correct way? give me any idea please. (This question is from elementary analysis textbook, so please assume that properties about $L_2$ convergence or almost everywhere convergence is not known)

Answer 1

Assume there is some point $x_0$ such that $f(x_0)\neq g(x_0)$ then there exist some open interval $I$ such that for all $x\in I : g(x)\neq f(x)$ by continuity of $f$ and $g$. Take a closed interval $J\subset I$ (Note that the measure of $J$ is strictly positive!). So there is some $A>0$ such that $A<|f(x)-g(x)|$ for all $x\in J$ (why?).

It appears to be given that $f_k \to g$ in $L^2$, hence there is a subsequence $f_{k_j}$ such that $f_{k_j} \to g$ almost everywhere in $[a,b]$.

Take $x\in J$ such that $f_{k_j}(x)\to g(x)$. There must exist such $x$ otherwise we could have that there is some non null set $J$ such that $f_{k_j}$ does not converge to $g$ for all $x\in J$. But that makes a contradiciton since $f_{k_j}\to g$ almost every where. After taking such $x$ we know that there exist some $N$ such that for all $k_j>N$ we have: \begin{align} |f_{k_j}(x)-g(x)| < \frac{A}{2} \ \ \ \text{ and }\ \ \ |f_{k_j}(x)-f(x)| < \frac{A}{2} \end{align}

So we have $x\in J$ and: \begin{align} A<|f(x)-g(x)|\leq |f_{k_j}(x)-g(x)| + |f_{k_j}(x)-f(x)| <\frac{A}{2}+\frac{A}{2}=A \end{align} We see that $A<A$ hence a contradiction.

So $f=g$.