The Integral of a Bounded, Finitely Supported, Measurable Function

Assume that $m(E) < \infty$ and let $\left\{ f_{n}: E \rightarrow \mathbb {R}\right\}$ be a sequence of bounded, measurable functions. Without referring to the Bounded Convergence Theorem or Egoroff's Theorem, give a direct proof that if $\left\{ f_{n}\right\} \rightarrow f$ uniformly on $E$, then $\lim_{n \rightarrow \infty } \int_{E} f_{n} d m = \int_{E} f d m$.

Let $m(E) < \infty$ and $\left\{ E_{n}\right\}_{n = 1}^{\infty }$ be a measurable partition of $E$. Just using the linearity of integration and the Bounded Convergence Theorem, show that $m(E) = \sum_{n = 1}^{\infty } m\left(E_{n}\right)$.

Let $f: E \rightarrow \mathbb {R}$ be bounded, finitely supported and measurable. For a measurable subset $A$ of $E$, show that $\left.\int_{A} f\right|_{A} d m = \int_{E} f \cdot \chi_{A} d m$.

Does the Bounded Convergence Theorem hold for the Riemann integral?

Let $f: E \rightarrow \mathbb {R}$ and $g: E \rightarrow \mathbb {R}$ be bounded, finitely supported and measurable, and assume that $f = g$ almost everywhere on $E$. Show that $\int_{E} f d m = \int_{E} g d m$.

Prove Theorem 2.

Theorem 2: If $f: \left[a, b\right] \rightarrow \mathbb {R}$ is a bounded, Riemann integrable function, then it is Lebesgue integrable and the two integrals are equal.