Proof of Bounded Measurable Functions on a Finite Measure Space Are Integrable

The constant function. Let $\mathbf{1}_{X}$ be the indicator function of $X$, so $\mathbf{1}_{X}(x)=1$ for every $x\in X$. By Basic Properties of the Integral of Nonnegative Measurable Maps (ii), $\mathbf{1}_{X}\in\mathcal{M}^{+}$ and

$$\int_{X}\mathbf{1}_{X}\,d\mu=\mu(X)<\infty,$$

so $\mu(X)$ is a real number (Small Facts about [0,∞]: Zero, Finiteness and Sums of Zeros (ii)). As a function into $\mathbb{R}$ with values $\ge0$, $\mathbf{1}_{X}$ is measurable (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (vii)), so $\mathbf{1}_{X}\in\mathcal{L}^{1}(\mu)$ and its integral as an element of $\mathcal{L}^{1}(\mu)$ is $\mu(X)$ (Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi)). Let $k:X\to\mathbb{R}$ be the constant function with value $B$ (Pointwise Operations on Real-Valued Maps (ii)). Then $k(x)=B\cdot1+0\cdot1$ for every $x\in X$, so $k=B\mathbf{1}_{X}+0\,\mathbf{1}_{X}$ (Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (v), Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (vii)), and by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii), $k\in\mathcal{L}^{1}(\mu)$ with

$$\int_{X}k\,d\mu=B\,\mu(X)+0\cdot\mu(X)=B\,\mu(X).$$

Proof of (i). It remains to show $f\in\mathcal{L}^{1}(\mu)$. The function $f$ is measurable, $k\in\mathcal{L}^{1}(\mu)$ and $|f(x)|\le B=k(x)$ for every $x\in X$, so $f\in\mathcal{L}^{1}(\mu)$ by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (v).

Proof of (ii). For every $x\in X$, $(f+B)(x)=1\cdot f(x)+1\cdot k(x)$, so $f+B=1f+1k$ (Arithmetic and Order Rules in the Real Numbers, Stated for ℝ (v)). By (i) and Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vii), $f+B\in\mathcal{L}^{1}(\mu)$ and, for the integral as an element of $\mathcal{L}^{1}(\mu)$,

$$\int_{X}(f+B)\,d\mu=\int_{X}f\,d\mu+\int_{X}k\,d\mu=\int_{X}f\,d\mu+B\,\mu(X).$$

The function $f+B$ is measurable, as $f$ and $k$ are (Measurable Maps: Generators, Composition, Continuous Maps, Restrictions, and Criteria for Real- and [0,∞]-Valued Maps (iii), Arithmetic of Measurable Real- and [0,∞]-Valued Maps (i)), and $f(x)+B\ge f(x)+|f(x)|\ge0$ for every $x\in X$, as $-|f(x)|\le f(x)$ (Rules of Arithmetic and Order in an Ordered Field (x), Rules of Arithmetic and Order in an Ordered Field (iv)). Hence, by Linearity and Monotonicity of the Lebesgue Integral, and the Integrability Criterion (vi), $f+B\in\mathcal{M}^{+}$, and, as $f+B\in\mathcal{L}^{1}(\mu)$, its integral as an element of $\mathcal{M}^{+}$ is finite and equals its integral as an element of $\mathcal{L}^{1}(\mu)$.