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If the derivatives exist for $x \in [0,1],$ then $f$ and $g$ must be continuous on some open set $A \subset [0,1].$ If $f \neq g$ on $A$ but $f' = g'$ a.e on $A$, then by integrating $f_1 = g_1 + C$, for some constant $C$ locally on open set $A$. However, because all the functions are continuous (locally on A) $f = g + C$ everywhere on $A$. It's enough to consider the problem locally. So the answer is $0.$

EDIT:(I now try to use FTC locally)

If the derivatives exist for $x \in [0,1],$ then $f$ and $g$ must be continuous on some open set $A \subset [0,1].$ If $f \neq g$ on $A$ but $f' = g'$ a.e on $A$, then by integrating $f_1 = g_1 + C$, for some constant $C$ locally on open set $A$. However, because all the functions are continuous (locally on A) $f = g + C$ everywhere on $A$. Moreover, it's enough to consider the problem locally, if the derivatives exist. So the answer is $0,$ if the derivatives exist.

EDIT:(I now try to use FTC locally)

If the derivatives exist for $x \in [0,1],$ then $f$ and $g$ must be continuous on some open set $A \subset [0,1].$ If $f \neq g$ on $A$ but $f' = g'$ on $A$, then by integrating $f = g + C$ a.e, for some constant $C$ locally on open set $A$. However, because both functions are continuous (locally on A) $f = g + C$ everywhere on $A$. So the answer is $0.$

EDIT:(I now try to use FTC locally)

If the derivatives exist for $x \in [0,1],$ then $f$ and $g$ must be continuous on some open set $A \subset [0,1].$ If $f \neq g$ on $A$ but $f' = g'$ a.e on $A$, then by integrating $f_1 = g_1 + C$, for some constant $C$ locally on open set $A$. However, because all the functions are continuous (locally on A) $f = g + C$ everywhere on $A$. It's enough to consider the problem locally. So the answer is $0.$

EDIT:(I now try to use FTC locally)

If $f \neq g$ on $A$ but $f' = g'$ a.e., then by integrating $f = g + C$ a.e, for some constant $C$. However, because both functions are continuous $f = g + C$ everywhere. So the answer is $0$.

If the derivatives exist for $x \in [0,1],$ then $f$ and $g$ must be continuous on some open set $A \subset [0,1].$ If $f \neq g$ on $A$ but $f' = g'$ on $A$, then by integrating $f = g + C$ a.e, for some constant $C$ locally on open set $A$. However, because both functions are continuous (locally on A) $f = g + C$ everywhere on $A$. So the answer is $0.$

EDIT:(I now try to use FTC locally)