Proof of Linearity of the Lebesgue Integral over a Finite Sum of Integrable Functions

Each result cited is universally quantified over the data in its own statement, and is applied here to the data named in the statement above. Let $S$ denote the successor map of Natural Numbers. We use below without further comment the identities $1\cdot t=t$ and $t+0=t$, valid for every real number $t$ by the axioms of Field, and the identity $0\cdot t=0$, which is claim 1 of Zero Products and Elementary Identities in a Field. An integrable map is measurable by Integrable Function and the Lebesgue Integral, so claim 1 follows once $g$ is shown to be integrable.

Notation. For $j\in[m]$ let $g_{j}:X\to\mathbb{R}$ be the map given by $g_{j}(x)=\sum_{k=1}^{j}c_{k}f_{k}(x)$, the finite sum being formed from the map $k\mapsto c_{k}f_{k}(x)$ on $[m]$ for each fixed $x$, and put $\Sigma_{j}=\sum_{k=1}^{j}c_{k}\int_{X}f_{k}\,d\mu$, formed from the map $k\mapsto c_{k}\int_{X}f_{k}\,d\mu$ on $[m]$. Thus $g_{m}=g$ and $\Sigma_{m}$ is the right-hand side of claim 2.

By Finite Sum Notation in a Field we have, for every $x\in X$,

and likewise $\Sigma_{1}=c_{1}\int_{X}f_{1}\,d\mu$ and $\Sigma_{S(j)}=\Sigma_{j}+c_{S(j)}\int_{X}f_{S(j)}\,d\mu$ whenever $S(j)\in[m]$.

We also record that if $S(j)\in[m]$ for a natural number $j$, then $j\in[m]$: indeed $S(j)\le m$ by Initial Segment of the Natural Numbers, while $j<S(j)$ and hence $j\le S(j)$ by claims 5 and 1 of Properties of the Order on the Natural Numbers, so $j\le m$ by claim 1 of that lemma.

The induction. For a natural number $j$ let $P(j)$ assert: if $j\in[m]$, then $g_{j}$ is integrable and $\int_{X}g_{j}\,d\mu=\Sigma_{j}$. The assertion is vacuously true when $j\notin[m]$. We prove $P(j)$ for every natural number $j$ by induction; taking $j=m$, which lies in $[m]$ by claim 1 of Properties of the Order on the Natural Numbers, then gives both claims.

Base. Suppose $1\in[m]$. The maps $f_{1}$ and $f_{1}$ are integrable and the reals $c_{1}$ and $0$ are given, so claim 2 of Linearity and Monotonicity of the Lebesgue Integral, applied with $f=f_{1}$, $g=f_{1}$, $a=c_{1}$ and $b=0$, shows that the map $x\mapsto c_{1}f_{1}(x)+0\cdot f_{1}(x)$ is integrable with integral $c_{1}\int_{X}f_{1}\,d\mu+0\cdot\int_{X}f_{1}\,d\mu$. In the field $\mathbb{R}$ we have $0\cdot t=0$ and $s+0=s$ for all $s,t$, so that map is $g_{1}$ and that integral is $\Sigma_{1}$. Hence $P(1)$.

Step. Let $j$ be a natural number, assume $P(j)$, and suppose $S(j)\in[m]$. Then $j\in[m]$ by the fact recorded above, so $P(j)$ applies: $g_{j}$ is integrable with $\int_{X}g_{j}\,d\mu=\Sigma_{j}$. Also $f_{S(j)}$ is integrable, by the hypothesis of the statement. Applying claim 2 of Linearity and Monotonicity of the Lebesgue Integral with $f=g_{j}$, $g=f_{S(j)}$, $a=1$ and $b=c_{S(j)}$, the map $x\mapsto 1\cdot g_{j}(x)+c_{S(j)}f_{S(j)}(x)$ is integrable with integral $1\cdot\int_{X}g_{j}\,d\mu+c_{S(j)}\int_{X}f_{S(j)}\,d\mu$. Since $1\cdot s=s$ in $\mathbb{R}$, that map is $g_{S(j)}$ by the recursion recorded above, and its integral is

Hence $P(S(j))$, and the induction is complete.