$X=\{x\in[a,b]; f(x)\neq g(x)\}$ has measure zero, then $\int f = \int g$

Question

Let $f,g:[a,b]\to\mathbb{R}$ integrables. If $X=\{x\in[a,b]; f(x)\neq g(x)\}$ has (Lebesgue) measure zero, then $\int_a^b f = \int_a^b g$.

I got a something and I need to check if it is correct:

Let $h=f-g$, then $h(x) \neq 0$ in $X$, and $h(x) = 0$ in $[a,b]-X$. Using Riemann sums, I need to show that $\forall \epsilon>0\; \exists \delta>0$ such that $|P|<\delta \implies \sum{|h(\xi_i)|\Delta t_i} < \epsilon$. Now, if the set $G_\delta = \{x\in[a,b]; |h(x)| \geq \delta\}$ has content zero for any $\delta>0$, then we are done. A zero content set $A$ is such that $\forall \epsilon>0$ there is $A\subset I_1\cup I_2\cup \cdots \cup I_n$ with $\sum |I_i| < \epsilon$, i.e. it is the finite version of measure zero. If a set $G_\delta \subset [a,b]$ has content zero, there exists a partition $P$ of $[a,b]$ such that the intervals containing points of $G_\delta$ are arbitrarily short. I will skip the finals details of this part, I am mainly interested in proving $G_\delta$ has content zero.

Now, to prove $G_\delta$ has content zero, I use the following theorem: A bounded function $h:[a,b]\to\mathbb{R}$ is integrable if and only if for any $\delta>0$, $E_\delta = \{x\in[a,b]; \omega_h(x) \geq \delta\}$ has content zero. Here, $\omega_h(x)$ denotes the oscillation of $h$ at $x$. Now if I can show that $|h(x)| \leq \omega(x)$, then $G_\delta \subset E_\delta$, and $G_\delta$ has content zero.

Defining $I_{x,\eta} = (x-\eta,x+\eta)\cap[a,b]$, we have $\omega_h(x) = \inf_\eta{\omega_h(I_{x,\eta})}$ where $\omega_h(I_{x,\eta}) = \sup \{|h(y)-h(z)|; y,z \in I_{x,\eta}\}$. Since $X$ has measure zero, any open set in $[a,b]$ contains points not in $X$, that is, points where $h$ is zero. Therefore $\omega_h(I_{x,\eta}) \geq |h(x)|$ for all $\eta>0$, then $\inf_\eta{\omega_h(I_{x,\eta})} = \omega_h(x) \geq |h(x)|$.

If you have other approaches, feel free to post them.

Answer 1

Consider $h(x) = f(x)- g(x)$. Then, as you said, $h(x) = 0, x \notin X$ and $h(x) \neq 0, x \in X$.

Then $\int f - \int g = \int (f - g) = \int h = \int_X h + \int_{[a, b] - X} h \leq max\{h(x) : x \in X\}Leb(X) + \int_{[a,b] - X} 0 = 0 + 0 = 0$.

Use a similar argument to show that $\int f - \int g \geq 0$.