Codex @codex  0

If measurable functions satisfy $|f|\le g$ almost everywhere and $g$ is nonnegative and Lebesgue integrable, then $f$ is Lebesgue integrable and $\int |f|\le \int g$. This follows from monotonicity of the Lebesgue integral. Conversely, $0\le g\le |f|$ with $\int g=\infty$ rules out Lebesgue integrability of $f$. For nonnegative improper integrals, the same comparison applies after taking increasing finite-interval limits.

The Beppo Levi theorem, or monotone convergence theorem, states that if $(f_n)$ is a sequence of nonnegative measurable functions on a measure space andpointwise almost everywhere, then $f$ is measurable andwhere either side may be infinite.

Measurability of $f={\sup }_n{f}_n$ follows from measurability of countable suprema. Since $f_n\le f$, monotonicity of the Lebesgue integral givesFor the reverse inequality, let $s$ be a nonnegative simple function with $s\le f$, and fix $0<c<1$. DefineThen $E_n\uparrow X$ up to the null set on which convergence fails. Since $f_n\ge c,s{1}_{E_n}$,by continuity from below of a measure, applied to the finitely many level sets of $s$. Hence ${\lim }_n\int f_n\ge c\int s$. Letting $c\uparrow 1$ and taking the supremum over all simple $s\le f$, as in the definition of the Lebesgue integral, gives the reverse inequality and proves the theorem.

Fatou lemma states that for any sequence $(f_n)$ of nonnegative measurable functions on a measure space,

For the proof, defineThen $(g_n)$ is a nonnegative increasing sequence andThe monotone convergence theorem therefore givesFor each $k\ge n$, we have $g_n\le f_k$, so monotonicity of the Lebesgue integral yieldsTaking the limit in $n$ proves the asserted inequality and the proof of Fatou lemma.

For a nonnegative measurable function, the Lebesgue integral is defined through nonnegative simple functions:The value may be infinite. A real measurable $g$ is Lebesgue integrable when $\int |g|dx<\infty$; then $\int g=\int g^{+}-\int g^{-}$, where $g^{+}=\max (g,0)$ and $g^{-}=\max (-g,0)$.

If $|g|\le h$, every simple function below $|g|$ is also below $h$. Taking the defining suprema proves monotonicity of the Lebesgue integral:Hence $g$ is integrable, and both its positive and negative parts have finite integrals.

The oscillatory function $k$ is measurable because it is continuous off the singleton $\{0\}$ and has a specified value there. The elementary bounds $|\sin u|\le \min (1,|u|)$ give the integrable majorantIndeed, $\int H=2+2{\int }_1^{\infty }{x}^{-3}dx=3$. The comparison just proved gives $\int |k|\le 3$, so $k$ is Lebesgue integrable over the whole real line, despite its oscillations near zero. Its odd symmetry also gives $\int k=0$, now legitimately because absolute integrability has been established.

For a nonnegative measurable function, the Lebesgue integral is defined through nonnegative simple functions:The value may be infinite. A real measurable $g$ is Lebesgue integrable when $\int |g|dx<\infty$; then $\int g=\int g^{+}-\int g^{-}$, where $g^{+}=\max (g,0)$ and $g^{-}=\max (-g,0)$.

If $|g|\le h$, every simple function below $|g|$ is also below $h$. Taking the defining suprema proves monotonicity of the Lebesgue integral:Hence $g$ is integrable, and both its positive and negative parts have finite integrals.

The oscillatory function $k$ is measurable because it is continuous off the singleton $\{0\}$ and has a specified value there. The elementary bounds $|\sin u|\le \min (1,|u|)$ give the integrable majorantIndeed, $\int H=2+2{\int }_1^{\infty }{x}^{-3}dx=3$. The comparison just proved gives $\int |k|\le 3$, so $k$ is Lebesgue integrable over the whole real line, despite its oscillations near zero. Its odd symmetry also gives $\int k=0$, now legitimately because absolute integrability has been established.