if $\{x| f(x) \neq g(x))\}$ has zero measure then $\int f = \int g$.

Question

Let $f,g$ be integrable functions in a closed box $B \subset \mathbb R^n$. Suppose that the set $A = \{x \in B| f(x) \neq g(x)\}$ is negligible. Show that $\int_Bf =\int_B g$

What I tried:

Notice that $B = A \cup (A^c \cap B)$ and that $A \cap (A^c \cap B) = \emptyset$ obviously.

Then $\int_Bf - \int_B g = \int_B f-g = \int_{A}f-g+\int_{A^c \cap B}f-g = \int_{A}f-g$

Now see that $0 = v(A)\inf_{x \in A}(f-g)(x) \leq \int_{A}f-g \leq v(A) \sup_{x \in A}(f-g)(x) =0$ because $v(A) = 0$ since it's negligible, and $f,g$ are integrable on $B$ so we know they are bounded.

Where is my issue:

I'm unsure about

$\int_B f-g = \int_{A}f-g+\int_{A^c \cap B}f-g$, I realize it's true when $A,A^c \cap B$ are boxes. If I split a big box into two smaller non intersecting boxes, then the integral on the big box is equal to the sum of the integrals of the two smaller boxes, but is it also true for general shapes?

Is there a way to prove this without resorting to integrals over non boxes?

Answer 1

We can also prove this in the framework of Riemann integration.

Take $h = f- g$ and

$$A = \{x \in B \,|\, f(x) \neq g(x) \} = \{ x \in B \,|\, h(x) \neq 0\}$$ where $A$ has zero measure.

If $P$ is any partition of $B$ into sub-rectangles (boxes) then any such sub-rectangle $R$ has non-zero content and so must include points $y$ not in A such that $h(y) = 0$.

Thus, $\inf_{x \in R}h(x) \leqslant 0 \leqslant \sup_{x \in R} h(x)$ for each $R \in P$ and the lower and upper Darboux sums satisfy

$$L(P,h) \leqslant 0 \leqslant U(P,h), \\ \implies L(P,h) \leqslant \sup_P L(P,h) \leqslant 0 \leqslant \inf_P \,U(P,h) \leqslant U(P,h)$$

However, we are given that $f$ and $g$, and , hence, $h$ are integrable.

Thus,

$$\int_B h = \sup_P L(P,h) \geqslant 0 , \quad \int_B h = \inf_P \,L(P,h) \leqslant 0,$$

which implies

$$\int_Bf - \int_B g = \int_Bh = 0$$

Answer 2

Since $f$ and $g$ are integrable, so is $f-g$, so $A=(f-g)^{-1}(\mathbb{R}\setminus\{0\})$ is a measurable set. Hence also $A^c\cap B$ is measurable.

Then use the fact that for any integrable function $f$ and partitioning measurable sets $A$ and $B$ we have $f=\chi_Af+\chi_Bf$ and hence $\int f=\int\chi_Af+\chi_Bf=\int\chi_Af+\int\chi_Bf=\int_Af+\int_Bf$.

Here $\chi_A$ denotes the indicator function, which is $1$ in $A$ and $0$ elsewhere.