Proof of Integrals of Functions Vanishing or Agreeing off a Null Set on a Compact Interval

Throughout write $\int f$ for $\int_{[a,b]}f\,d\lambda_{[a,b]}$, write $\mathbf{0}$ for the function on $[a,b]$ whose value is $0$ at every point, and for $E\subseteq[a,b]$ write $\mathbf{1}_E$ for the function equal to $1$ on $E$ and to $0$ elsewhere.

The integral of $\mathbf{0}$. The function $\mathbf{0}$ is measurable and nonnegative. By the definition of the integral of a nonnegative measurable function, $\int\mathbf{0}$ is the supremum of the numbers $\int s$ over the simple functions $s$ on $[a,b]$ with $0\le s(t)\le\mathbf{0}(t)$ for every $t$. The only such $s$ is $\mathbf{0}$ itself, whose standard representation takes the single value $0$ on $[a,b]$, so that its integral is $0\cdot\lambda_{[a,b]}([a,b])=0$. The supremum of the set whose only element is $0$ is $0$, so $\int\mathbf{0}=0$.

Claim 1. Put $D=[a,b]\setminus N$. Then $D\in\mathcal{B}_{[a,b]}$, because $\mathcal{B}_{[a,b]}$ is a $\sigma$-algebra on $[a,b]$ by claim 1 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval and hence closed under complements relative to $[a,b]$, and $\lambda_{[a,b]}([a,b]\setminus D)=\lambda_{[a,b]}(N)=0$; thus $D$ is co-null in the sense of claim 6 of that lemma.

The absolute value is continuous from the real line to itself, hence measurable with respect to the Borel $\sigma$-algebra of the real line by claim 3 of Borel Measurability and Bounded Integration on a Metric Space together with claim 2 there, which identifies the Borel $\sigma$-algebra of the real line as a metric space with $\mathcal{B}(\mathbb{R})$. Therefore the composition $|h|$, whose value at $t$ is the absolute value of $h(t)$, is measurable by claim 4 of Borel Measurability and Bounded Integration on a Metric Space, and $|h|\ge0$ by claim 1 of Properties of the Absolute Value in an Ordered Field.

Since $h(t)=0$ for every $t\in D$, the function $|h|\,\mathbf{1}_D$ has value $0$ at every point of $[a,b]$: at $t\in D$ because $|h(t)|=0$, and at $t\notin D$ because $\mathbf{1}_D(t)=0$. That is, $|h|\,\mathbf{1}_D=\mathbf{0}$. Applying claim 6 of Restricted Lebesgue Measure and Integral Toolkit on a Compact Interval to the nonnegative measurable function $|h|$ and the co-null set $D$ gives

In particular $\int|h|$ is finite, so $h$ is integrable by the definition of integrability.

By claim 3 of Properties of the Absolute Value in an Ordered Field we have $-|h(t)|\le h(t)\le|h(t)|$ for every $t$, so $-|h|\le h\le|h|$ pointwise. The functions $h$, $|h|$ and $-|h|$ are all integrable, the last by the linearity in claim 2 of Linearity and Monotonicity of the Lebesgue Integral. The monotonicity in that same claim gives

and the linearity there gives $\int(-|h|)=-\int|h|$. Since $\int|h|=0$ and $-0=0$ by claim 4 of Additive Cancellation and Elementary Additive Identities in a Field, these two inequalities read $\int h\le0$ and $0\le\int h$. Two real numbers each at most the other are equal, so $\int h=0$.

Claim 2. The function $h=f-g$ is integrable by the linearity in claim 2 of Linearity and Monotonicity of the Lebesgue Integral, and in particular measurable. For $t\in[a,b]\setminus N$ we have $f(t)=g(t)$ and hence $h(t)=0$, by claim 3 of Additive Cancellation and Elementary Additive Identities in a Field. So claim 1 applies to $h$ and gives $\int(f-g)=0$. By the linearity in claim 2 of Linearity and Monotonicity of the Lebesgue Integral again, $\int f-\int g=\int(f-g)=0$, and therefore $\int f=\int g$.