The Lebesgue Integral and the Convergence Theorems — Examples

For each $n$, put [L1, L2, construct] $s_n:={\sum }_{k<n}\min \{f(k),n\}{\chi }_{\{k\}}\mathrm{.}$ Then $s_n$ is finite-valued and simple, $s_n\uparrow f$, and [L3] gives $\int s_{n}d\mathrm{\#}={\sum }_{k<n}\min \{f(k),n\}\mathrm{\#}(\{k\})={\sum }_{k<n}\min \{f(k),n\}$ because each singleton has counting measure $1$ by [L1].

Applying [L3] to $s_n\uparrow f$ gives [step 1.1, L3] ∎ $\int fd\mathrm{\#}={\lim }_n{\sum }_{k<n}\min \{f(k),n\}\mathrm{.}$ The diagonal truncated sums increase to the nonnegative extended series ${\sum }_{k\in \mathbb{N}}f(k)$: each is at most that series, while every fixed finite partial sum is approached from below as $n\to \mathrm{\infty }$. This proves the displayed identity.

By [L1], the restriction of ${\mathbf{1}}_{\mathbb{Q}}$ to $[0,1]$ is positive at every rational point, hence on a dense subset of $[0,1]$.

By [L2], the set $\mathbb{Q}\cap [0,1]$ on which ${\mathbf{1}}_{\mathbb{Q}}$ is positive is null. Therefore ${\mathbf{1}}_{\mathbb{Q}}=0$ almost everywhere on $[0,1]$, so [L3] gives.

Let $f_n:=e^{-x}{\chi }_{[0,n]}$. Then $f_n\uparrow f$. For each $k\ge 0$ and [L1, L2, construct] $x\in [k,k+1]$, one has $e^{-x}\le e^{-k}$, so ${\int }_{[k,k+1]}{f}_{n}d\lambda \le e^{-k}\lambda ([k,k+1])=e^{-k}$ by [L2].

Additivity [L4] therefore gives [step 1.1, L1, L3, L4] ∎ $\int f_{n}d\lambda \le {\sum }_{k<n}{e}^{-k},$ and the right-hand side is bounded independently of $n$ by [L3]. Passing to the limit with [L1] shows ${\int }_0^{\mathrm{\infty }}{e}^{-x}d\lambda <+\mathrm{\infty }$.

On the dyadic interval $I_k:=(2^{-k-1},2^{-k}]$, one has $f(x)\le 2^{(k+1)\mathrm{/}2}$ and $\lambda (I_k)=2^{-k-1}$.

The partial sums of the integrals over ${\bigcup }_{k<n}{I}_k=(2^{-n},1]$ are bounded by ${\sum }_{k<n}{2}^{-(k+1)\mathrm{/}2}$.

If $s={\sum }_j{c}_j{\chi }_{E_j}$ is simple, then [L1, given, algebra] $\int sd{\delta }_{x_0}={\sum }_j{c}_j{\delta }_{x_0}(E_j)=s(x_0),$ because exactly one cell containing $x_0$ contributes.

For nonnegative measurable $f$, choose simple $s_n\uparrow f$ by [L2]. Then [step 1.1, L2, L3] ∎ $\int fd{\delta }_{x_0}={\lim }_n\int s_{n}d{\delta }_{x_0}={\lim }_n{s}_n(x_0)=f(x_0)\mathrm{.}$ Apply this to the positive/negative parts and to the real/imaginary parts to obtain the same formula for real and complex integrable $f$ by [L3].

Fix a compact interval $[a,b]\subset (0,\mathrm{\infty })$. For $f(x,t):=e^{-tx}\sin x,$ one has $\frac{\mathrm{\partial }f}{\mathrm{\partial }t}(x,t)=-x{e}^{-tx}\sin x,$ so $\mid \frac{\mathrm{\partial }f}{\mathrm{\partial }t}(x,t)\mid \le x{e}^{-ax}(t\in [a,b])\mathrm{.}$

The function $x\mapsto x{e}^{-ax}$ is integrable on $[0,\mathrm{\infty })$, so [L1] applies on every compact parameter interval and yields $F^{\mathrm{\prime }}(t)=-{\int }_0^{\mathrm{\infty }}x{e}^{-tx}\sin xdx\mathrm{.}$ This is exactly the advertised differentiation step.

Put a discrete probability measure on $\{1,\dots ,m\}$ by[L1, construct] $\mathbb{P}(\{j\})={\lambda }_j$, let $f(j)=a_j$, and choose the convex function $\phi (x)=-\log x$ on $(0,\mathrm{\infty })$. Applying [L1] gives $-\log \text{\hspace{0.17em}⁣}\left({\sum }_{j=1}^m{\lambda }_j{a}_j\right)\le -{\sum }_{j=1}^m{\lambda }_j\log a_j\mathrm{.}$

Multiply by $-1$ and exponentiate to obtain [step 1.1, algebra] ∎ ${\prod }_{j=1}^m{a}_j^{{\lambda }_j}\le {\sum }_{j=1}^m{\lambda }_j{a}_j,$ the weighted AM-GM inequality.

The sequence $f_n$ converges pointwise almost everywhere to $0$, but ${\int }_0^1{f}_{n}d\lambda =(n+1)\lambda ((0,1\mathrm{/}(n+1)))=1$ for every $n$.

Therefore [step 1.1, L1, L2] ∎ $\int {\mathrm{lim\hspace{0.17em}inf}}_n{f}_{n}d\lambda =0<1={\mathrm{lim\hspace{0.17em}inf}}_n\int f_{n}d\lambda ,$ so Fatou is strict, and the unchanged integral also shows that no dominated convergence conclusion can hold. This is exactly the hypothesis loss recorded in [L1] and [L2].

Each $f_n$ has finite support, so it is integrable on [L1, L2, given] $(\mathbb{N},\mathcal{P}(\mathbb{N}),\mathrm{\#})$.

For every $k\in \mathbb{N}$, one has $f_n(k)\to 1$. The pointwise limit is the [step 1.1, L1, L2, algebra] ∎ constant function $1$, whose integral under counting measure is $+\mathrm{\infty }$, so it is not integrable by [L2]. Thus the Statement is false.

For each fixed $x\in \mathbb{R}$, the value $f_n(x)$ is eventually $0$, so $f_n(x)\to 0$.

Yet $\int f_{n}d\lambda =1$ for every $n$. Thus the mass has escaped to infinity instead of disappearing, and the Statement is false. This is the strict case already permitted by [L1].

Since $0\le f_n\le 1\mathrm{/}(n+1)$, the sequence $(f_n)$ converges uniformly to $0$ on $\mathbb{R}$.

Nevertheless, ${\int }_{\mathbb{R}}{f}_{n}d\lambda =(n+1{)}^{-1}\lambda ([0,n+1])=1$ for every $n$, so the integrals do not converge to $0$. This refutes the Statement and shows why [L1] needs finite total measure.

The functions $f_n$ decrease pointwise to $0$.

But $\int f_{n}d\lambda =+\mathrm{\infty }$ for every $n$, while $\int 0d\lambda =0$. [step 1.1, L1, algebra] ∎ So the displayed conclusion fails, confirming the directionality recorded in [L1].

The sum $f+g$ is the zero function, so its integral is $0$.

But $f$ has integral $+\mathrm{\infty }$, while $g$ would have to contribute [step 1.1, L1, algebra] ∎ $-\mathrm{\infty }$ for any linear identity to hold. Thus $\int (f+g)d\lambda =0\mathrm{\ne }+\mathrm{\infty }+(-\mathrm{\infty }),$ so the Statement is false and [L1] cannot be widened beyond $L^1$.

Under counting measure,[L2, given, algebra] $\int fd\mathrm{\#}=2,\int \phi (f)d\mathrm{\#}=2.$

Hence [step 1.1, L1] ∎ $\phi \text{\hspace{0.17em}⁣}(\int fd\mathrm{\#})=4>2=\int \phi (f)d\mathrm{\#}\mathrm{.}$ So Jensen fails on this infinite measure space, exactly as warned by [L1].