Let $f,g : [0,T] \to X$ be continuous $X$-valued functions for Banach Space $(X,||\, \cdot\,||)$. Suppose that $f = g$ almost everywhere in $C([0,T],X)$, then $f = g$ in $C([0,T],X)$.
This is my attempt so far :
First, $f = g$ almost everywhere in $C([0,T],X)$ means $\forall t \in [0,T]\backslash M, f(t) = g(t)$ for some $M$ measure zero set. Therefore, our claim is $M \neq \emptyset$.
By contradiction, suppose there exists $t_{1} \in M\subset [0,T]$ such that $f(t_{1}) \neq g(t_{1})$. Then, we know that $||f(t_{1}) – g(t_{1})|| \neq 0$.
Furthermore, from our assumption, we know that $\forall t \in [0,T]\backslash M, ||f(t) – g(t)|| = 0$.
However, since $f$ and $g$ are in $C([0,T],X)$, then $\forall \varepsilon > 0, \exists \delta > 0$ such that $\forall t \in (t_{1}-\delta, t_{1}+\delta), f(t)\neq g(t)$. Therefore, $\forall t \in (t_{1}-\delta, t_{1}+\delta), ||f(t) – g(t)||\neq 0$. Fix $\varepsilon = \varepsilon_{1}$ to obtain $\delta = \delta_{1}$. Then, $(t_{1}-\delta_{1},t_{1}+\delta_{1}) \subset M$ which means measure of $M$ is not zero. This is a contradiction and therefore $M = \emptyset$ which implies $\forall t \in [0,T], f(t) = g(t)$ in $C([0,T],X)$.
So my question is where did I make any mistake? Also, I want to make this argument more rigorous so I would like to ensure $f(t) \neq g(t)$ in $(t_{1} – \delta_{1}, t_{1} + \delta_{1}$). Is my reasoning enough to ensure that? I tried to use $\varepsilon_{1} = \frac{1}{2}||f(t_{1}) – g(t_{1})||$ but I am unsure whether I am correct or not.
Any help is very much appreciated! Thank you!
Best Answer
Your argument is generally correct. A few comments:
I don't see what the role of $\varepsilon_1$ and $\delta_1$ is supposed to be. Just use $\varepsilon$ and $\delta$.
By working with $h=f-g$, you may work with a single continuous function, proving that if $h=0$ a.e., then $h=0$.
As you mention at the end, you want to take a concrete $\varepsilon$. Taking $\varepsilon=\frac12\,\|h(t_1)\|$ is a good choice, because then you may use the reverse triangle inequality to get, for $t\in (t_1-\delta,t_1+\delta)$, \begin{align} \|h(t)\|&=\|h(t)-h(t_1)+h(t_1)\|\geq\|h(t_1)\|-\|h(t)-h(t_1)\|\\ \ \\ &\geq\|h(t_1)\|-\frac12\|h(t_1)\|=\frac12\,\|h(t_1)\|>0. \end{align}