Let $0<|{x}_{0}-x|<{\displaystyle \frac{\delta}{2}.}$ Further, let $0<{y}_{0}-y<{\displaystyle \frac{\delta}{2},}$, and (*x*_{0},*y*_{0}) be a common μ-generalized Lebesgue point of the functions $f\in {L}_{p}({}^{2})(1p<\mathrm{\infty})\phantom{\rule{thinmathspace}{0ex}}\mathrm{a}\mathrm{n}\mathrm{d}\phantom{\rule{thinmathspace}{0ex}}\phi .$ The proof of theorem will be given for the case 1 < p < ∞. The proof for the case *p* = 1 is similar. Now, set $I(x,y,\lambda )=|{T}_{\lambda}(f;x,y)-f({x}_{0},{y}_{0})|.$ Using condition (c) , we obtain
$$I(x,y,\lambda )=\left|\underset{{\mathbb{R}}^{2}}{\iint}{K}_{\lambda}(t-x,s-y,f(t,s))dsdt-f({x}_{0},{y}_{0})\right|=\left|\underset{{\mathbb{R}}^{2}}{\iint}{K}_{\lambda}(t-x,s-y,f(t,s))dsdt-\underset{{\mathbb{R}}^{2}}{\iint}{K}_{\lambda}(t-x,s-y,\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\phi (t,s))dsdt+\underset{{\mathbb{R}}^{2}}{\iint}{K}_{\lambda}(t-x,s-y,\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0;}{y}_{0})}\phi (t,s))dsdt-f({x}_{0},{y}_{0})\right|$$

Using condition (*b*), it is easy to see that the following inequality holds:
$$I(x,y,\lambda )\le \underset{{\mathbb{R}}^{\mathbb{2}}}{\iint}\left|\frac{f(t,s)}{(t,s)}-\frac{f({x}_{0},{y}_{0})}{({x}_{0},{y}_{0})}\right|\phi (t,s){L}_{\lambda}(t-x,s-y)dsdt+\left|\underset{{\mathbb{R}}^{2}}{\iint}{K}_{\lambda}(t-x,s-y,\frac{f({x}_{0},{y}_{0})}{({x}_{0},{y}_{0})}\phi (t,s))dsdt-f({x}_{0},{y}_{0})\right|.$$

Since whenever *m*,*n* being positive numbers the inequality $(m+n{)}^{p}\le {2}^{p}({m}^{p}+{n}^{p})$ holds (see, e.g., 38) we have
$$[I(x,y,\lambda ){]}^{p}\le {2}^{p}{\left(\underset{{\mathbb{R}}^{2}}{\iint}\left|\frac{f(t,s)}{\phi (t,s)}-\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\right|\phi (t,s){L}_{\lambda}(t-x,s-y)dsdt\right)}^{p}+{2}^{p}{\left|\underset{{\mathbb{R}}^{2}}{\iint}{K}_{\lambda}(t-x,s-y,\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\phi (t,s))dsdt-f({x}_{0},{y}_{0})\right|}^{p}.$$

Now, applying Hölder's inequality (see, e.g., 38) to the first integral of the resulting inequality, we have
$$[I(x,y,\lambda ){]}^{p}\le {2}^{p}\beta (x,y,\lambda )\underset{{\mathbb{R}}^{2}}{\iint}{\left|\frac{f(t,s)}{\phi (t,s)}-\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\right|}^{p}\phi (t,s){L}_{\lambda}(t-x,s-y)dsdt+{2}^{p}{\left|\underset{{\mathbb{R}}^{2}}{\iint}{K}_{\lambda}(t-x,s-y,\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\phi (t,s))dsdt-f({x}_{0},{y}_{0})\right|}^{p},$$ where
$$\beta (x,y,\lambda )={\left(\underset{{\mathbb{R}}^{2}}{\iint}\phi (t,s){L}_{\lambda}(t-x,s-y)dsdt\right)}^{\frac{p}{q}}.$$

Moreover, the following inequality holds:
$$[I(x,y,\lambda ){]}^{p}\le {2}^{p}\beta (x,y,\lambda )\underset{{\mathbb{R}}^{2}\mathrm{\setminus}{B}_{\delta}}{\iint}{\left|\frac{f(t,s)}{\phi (t,s)}-\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\right|}^{p}\phi (t,s){L}_{\lambda}(t-x,s-y)dsdt$$
$$+{2}^{p}\beta (x,\phantom{\rule{thickmathspace}{0ex}}y,\phantom{\rule{thickmathspace}{0ex}}\lambda )\underset{{B}_{\delta}}{\iint}|\frac{f(t,s)}{\phi (t,s)}-\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}{|}^{p}\phi (t,\phantom{\rule{thickmathspace}{0ex}}s){L}_{\lambda}(t-x,\phantom{\rule{thickmathspace}{0ex}}s-y)dsdt+{2}^{p}|{\iint}_{{\mathbb{R}}^{2}}{K}_{\lambda}(t-x,s-y,\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\phi (t,s))dsdt-f({x}_{0},{y}_{0}){|}^{p}$$ where
$(t-{x}_{0}{)}^{2}+(s-{y}_{0}{)}^{2}<{\delta}^{2}\}.$

Now, applying the inequality given by
$(m+n{)}^{p}\underset{\_}{<}{2}^{p}({m}^{p}+{n}^{p})$
once more to the right hand side of the resulting inequality, we
obtain$$[I(x,y,\lambda ){]}^{p}\underset{\_}{<}{2}^{2p}\beta (x,y,\lambda )|\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}{|}^{p}{\int}_{{\mathrm{R}}^{2}}{\int}_{\mathrm{\setminus}{B}_{\delta}}\phi (t,s){L}_{\lambda}(t-x,s-y)dsdt+{2}^{2p}\beta (x,y,\lambda ){\int}_{{\mathrm{R}}^{2}}{\int}_{\mathrm{\setminus}{B}_{\delta}}|\frac{f(t,s)}{\phi (t,s)}{|}^{p}\phi (t,s){L}_{\lambda}(t-x,s-y)dsdt+{2}^{p}\beta (x,y,\lambda ){\int}_{B}{\int}_{\delta}|\frac{f(t,s)}{\phi (t,s)}-\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}{|}^{p}\phi (t,s){L}_{\lambda}(t-x,s-y)dsdt+{2}^{p}|{\iint}_{{\mathrm{R}}^{2}}{K}_{\lambda}(t-x,s-y,\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\phi (t,s))dsdt-f({x}_{0},{y}_{0}){|}^{p}$$

Recalling the initial assumptions
$0<|{y}_{0}-y|<{\displaystyle \frac{\delta}{2}}$, we
may define the following set as follows:$${N}_{\delta}=\{(x,y)\in {\mathbb{R}}^{2}:(x-{x}_{0}{)}^{2}+(y-{y}_{0}{)}^{2}<\frac{{\delta}^{2}}{2}\}.$$

Comparing geometric representations of the
sets${B}_{\delta}$ gives the inclusion relation such
that${\mathbb{R}}^{2}\mathrm{\setminus}{B}_{\delta}\subseteq {\mathbb{R}}^{2}\mathrm{\setminus}{A}_{\delta},$ where$${A}_{\delta}=\{(t,s)\in {B}_{\delta}:(t-x{)}^{2}+(s-y{)}^{2}<\frac{{\delta}^{2}}{2},(x,y)\in {N}_{\delta}\}.$$

In the light of these relations, we may write$$[I(x,y,\lambda ){]}^{p}\underset{\_}{<}{2}^{2p}\phi (x,y)\beta (x,y,\lambda )|\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}{|}^{p}\underset{{R}^{2}\mathrm{\setminus}{A}_{\delta}}{\iint}\phi (t-x,s-y){L}_{\lambda}(t-x,s-y)dsdt+{2}^{2p}\phi (x,y)\beta (x,y,\lambda )\underset{(t,s)\in {\mathrm{R}}^{2}\mathrm{\setminus}{A}_{\delta}}{sup}[\phi (t-x,s-y){L}_{\lambda}(t-x,s-y)]\parallel f{\parallel}_{{L}_{p}^{\phi}({\mathrm{R}}^{2})}^{p}+{2}^{p}\beta (x,y,\lambda )\underset{{B}_{\delta}}{\iint}|\frac{f(t,s)}{\phi (t,s)}-\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}{|}^{p}\phi (t,s){L}_{\lambda}(t-x,s-y)dsdts+{2}^{p}|\underset{{{\mathbb{R}}^{\mathbb{2}}}_{{}^{}}}{\iint}{K}_{\lambda}(t-x,s-y,\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\phi (t,s))dsdt-f({x}_{0},{y}_{0}){|}^{p}$$

Rearranging and rewriting the last inequality, we obtain $$[I(x,y,\lambda ){]}^{p}\underset{\_}{<}{2}^{2p}\phi (x,y)\beta (x,y,\lambda )|\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}{|}^{p}\underset{\sqrt{{u}^{2}+{v}^{2}}\frac{{\delta}^{2}}{2}}{\iint}\phi (u,\phantom{\rule{thickmathspace}{0ex}}v){L}_{\lambda}(u,\phantom{\rule{thickmathspace}{0ex}}v)dvdu$$
$$+{2}^{2p}\phi (x,\phantom{\rule{thickmathspace}{0ex}}y)\beta (x,\phantom{\rule{thickmathspace}{0ex}}y,\phantom{\rule{thickmathspace}{0ex}}\lambda )\phantom{\rule{thickmathspace}{0ex}}\underset{\sqrt{{u}^{2}+{v}^{2}}\u2a7e\frac{{\delta}^{2}}{2}}{sup}[\phi (u,\phantom{\rule{thickmathspace}{0ex}}v){L}_{\lambda}(u,\phantom{\rule{thickmathspace}{0ex}}v)]{f}_{{L}_{p}^{\phi}({\text{R}}^{2})}^{p}+{2}^{p}\beta (x,y,\lambda \underset{{B}_{\delta}}{\iint}{\left|\frac{f(t,s)}{\phi (t,s)}-\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\right|}^{p}\phi (t,s){L}_{\lambda}(t-x,s-y)dsdt+{2}^{p}{\left|{\iint}_{{\mathrm{R}}^{2}}{K}_{\lambda}(t-x,s-y,\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\phi (t,s))dsdt-f({x}_{0},{y}_{0})\right|}^{p}={I}_{1}(x,y,\lambda )+{I}_{2}(x,y,\lambda )+{2}^{p}\beta (x,y,\lambda ){I}_{3}(x,y,\lambda )+{I}_{4}(x,y,\lambda ).$$

Boundedness of the term $(x,y,\lambda )$ which follows from condition (*f*).On the other hand, ${I}_{4}(x,y,\lambda )\to 0$ by
condition(*c*).Lastly,
${I}_{2}(x,y,\lambda )\to 0$ by
conditions (*d*) and (*e*) ,
respectively.

Now, we may write the following inequality for the
integral ${I}_{3}(x,y,\lambda )$
$${I}_{3}(x,y,\lambda )=\underset{{B}_{\delta}}{\iint}{\left|\frac{f(t,s)}{\phi (t,s)}-\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\right|}^{p}\phi (t,s){L}_{\lambda}(t-x,s-y)dsdt\underset{\_}{<}\underset{(t,s)\in {Q}_{\delta}}{sup}\phi (t,s){\int}_{Q}{\int}_{\delta}{\left|\frac{f(t,s)}{\phi (t,s)}-\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\right|}^{p}{L}_{\lambda}(t-x,s-y)dsdt=\underset{(t,s)\in {Q}_{\delta}}{sup}\phi (t,s){I}_{31}(x,y,\lambda ),$$ where
${Q}_{\delta}=({x}_{0}-\delta ,{x}_{0}+\delta )\times ({y}_{0}-\delta ,{y}_{0}+\delta ).$Observe that *I*_{31}(*x,y,λ)* may be written in the following form:$${I}_{31}(x,y,\lambda )=\underset{{Q}_{\delta}}{\iint}{\left|\frac{f(t,s)}{\phi (t,s)}-\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\right|}^{p}{L}_{\lambda}(t-x,s-y)dsdt=\underset{{x}_{0}-\delta}{\overset{{x}_{0}+\delta y}{\int}}\underset{{y}_{0}-\delta}{\overset{{y}_{0}+\delta}{\int}}{\left|\frac{f(t,s)}{\phi (t,s)}-\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}+\frac{f(t,{y}_{0})}{\phi (t,{y}_{0})}-\frac{f(t,{y}_{0})}{\phi (t,{y}_{0})}\right|}^{p}\times {L}_{\lambda}(t-x,s-y)dsdt.$$

It is easy to see that the following inequality holds:$${I}_{31}(x,y,\lambda )\underset{\_}{<}{2}^{p}\underset{{x}_{0}-\delta}{\overset{x\mathrm{o}+\delta}{\int}}\left[\underset{\mathcal{y}0-\delta}{\overset{\mathcal{y}0+\delta}{\int}}{\left|\frac{f(t,s)}{\phi (t,s)}-\frac{f(t,{y}_{0})}{\phi (t,{y}_{0})}\right|}^{p}{L}_{\lambda}(t-x,s-y)ds\right]dt+{2}^{p}\underset{{y}_{0}-\delta}{\overset{\mathcal{y}0+\delta}{\int}}\left[\underset{x\mathrm{o}+\delta}{\overset{x\mathrm{o}+\delta}{\int}}{\left|\frac{f(t,{y}_{0})}{\phi (t,{y}_{0})}-\frac{f({x}_{0},{y}_{0})}{\phi ({x}_{0},{y}_{0})}\right|}^{p}{L}_{\lambda}(t-x,s-y)dt\right]ds={2}^{p}\left\{\underset{{x}_{0}-\delta}{\overset{{x}_{0}+\delta}{\int}}{i}_{\lambda}(x,y,t)dt+\underset{{y}_{0}-\delta}{\overset{\mathcal{y}0+\delta}{\int}}{j}_{\lambda}(x,y,s)ds\right\}.$$Now, we pass to the integral
${i}_{\lambda}(x,y,t)$.

This integral may be written in the following form: $${i}_{\lambda}(x,y,t)=\underset{\mathcal{y}0-\delta}{\overset{\mathcal{y}0+\delta}{\int}}{\left|\frac{f(t,s)}{\phi (t,s)}-\frac{f(t,{y}_{0})}{\phi (t,{y}_{0})}\right|}^{p}{L}_{\lambda}(t-x,s-y)ds=\left\{\underset{{y}_{0}-\delta}{\overset{\mathcal{y}0}{\int}}+\underset{\mathcal{y}0}{\overset{\mathcal{y}0+\delta}{\int}}\right\}{\left|\frac{f(t,s)}{\phi (t,s)}-\frac{f(t,{y}_{0})}{\phi (t,{y}_{0})}\right|}^{p}{L}_{\lambda}(t-x,s-y)ds={i}_{\lambda}^{1}(x,y,t)+{i}_{\lambda}^{2}(x,y,t).$$

Now, we consider the integral${i}_{\lambda}^{1}(x,y,t)$ From relation (5),
for every $\u03f5>0$
there exists a corresponding
number$\delta >0$
such that the expression $$\underset{\mathcal{Y}0-k}{\overset{\mathcal{Y}0}{\int}}{\left|\frac{f(t,s)}{\phi (t,s)}-\frac{f(t,{y}_{0})}{\phi (t,{y}_{0})}\right|}^{p}ds<{\u03f5}^{p}{\mu}_{2}(k)$$(9) holds for every
$0<k{\displaystyle \underset{\_}{<}\delta <min\{{\delta}_{0},{\delta}_{1}\}.}$

Define the function *F*(*t*,s) by
$$F(t,{\displaystyle s)=\underset{s}{\overset{\mathcal{Y}0}{\int}}{\left|\frac{f(t,w)}{\phi (t,w)}-\frac{f(t,{y}_{0})}{\phi (t,{y}_{0})}\right|}^{p}dw}$$10

From (10), we have
$${d}_{s}F(t,{\displaystyle s)=-{\left|\frac{f(t,s)}{\phi (t,s)}-\frac{f(t,{y}_{0})}{\phi (t,{y}_{0})}\right|}^{p}ds}$$11

From (9) and (10), for every
*s*satisfying$0<{y}_{0}-s{\displaystyle \underset{\_}{<}\delta <min\{{\delta}_{0},{\delta}_{1}\}}$ we have
$$|F(t,s)|\underset{\_}{<}{\u03f5}^{p}{\mu}_{2}({y}_{0}-s)$$12(12)for
any fixed $t\in \mathbb{R}$ By virtue of (10) and (11), we have
$${i}_{\lambda}^{1}(x,y,t)=(L)\underset{\mathcal{y}0-\delta}{\overset{\mathcal{y}0}{\int}}{\left|\frac{f(t,s)}{\phi (t,s)}-\frac{f(t,{y}_{0})}{\phi (t,{y}_{0})}\right|}^{p}{L}_{\lambda}(t-x,s-y)ds=(LS)\underset{{y}_{0}-\delta}{\overset{\mathcal{y}0}{\int}}{L}_{\lambda}(t-x,s-y){d}_{s}[-F(t,s)],$$where (*LS*) denotes Lebesgue-Stieltjes integral.

Using integration by parts and applying (12), we have the following inequality:$$|{i}_{\lambda}^{1}(x,y,t)|\underset{\_}{<}{\u03f5}^{p}{\mu}_{2}(\delta ){L}_{\lambda}(t-x,{y}_{0}-\delta -y)+{\u03f5}^{p}\underset{\mathcal{y}0-\delta}{\overset{\mathcal{y}0}{\int}}{\mu}_{2}({y}_{0}-s)|{d}_{s}{L}_{\lambda}(t-x,s-y)|.$$

It is easy to see that (for the similar situation, see [7, 9]) the following inequality holds:$$\left|{i}_{\lambda}^{1}(x,y,t)\right|\underset{\_}{<}{\u03f5}^{p}{\mu}_{2}(\delta ){L}_{\lambda}(t-x,{y}_{0}-\delta -y)+{\u03f5}^{p}\underset{\mathcal{y}0-\delta}{\overset{\mathcal{y}0}{\int}}{\mu}_{2}({y}_{0}-s-y)\left|{d}_{s}\left[\underset{{y}_{0}-y-\delta}{\overset{s}{V}}(t-x,u)\right]\right|.$$

Applying integration by parts to the right hand side of the last inequality, we have the following expression:$$|{i}_{\lambda}^{1}(x,y,t)|\underset{\_}{<}-{\u03f5}^{p}\underset{{y}_{0}-y-\delta}{\overset{{y}_{0}-y}{\int}}\left[\underset{{y}_{0}-y-\delta}{\overset{s}{V}}{L}_{\lambda}(t-x,u)\right]{{\left\{{\mu}_{2}({y}_{0}-s-y)\right\}}^{\mathrm{\prime}}}_{s}ds.$$

Remaining variational operations are evaluated using condition (*g*). Hence, we obtain
$$|{i}_{\lambda}^{1}(x,{\displaystyle y,t)|\underset{\_}{<}{\u03f5}^{p}\underset{{y}_{0}-\delta}{\overset{\mathcal{Y}0}{\int}}{L}_{\lambda}(t-x,s-y)\left|\{{\mu}_{2}({y}_{0}-s){\}}_{s}^{\prime}\right|ds+2{L}_{\lambda}(t-x,0){\mu}_{2}({y}_{0}-y)}$$13

Using preceding method, we can estimate the integral
${i}_{\lambda}^{2}(x,y,t)$as follows:
$$|{i}_{\lambda}^{2}(x,{\displaystyle y,t)|\underset{\_}{<}{\u03f5}^{p}\underset{\mathcal{y}0}{\overset{\mathcal{y}0+\delta}{\int}}{L}_{\lambda}(t-x,s-y)|\{{\mu}_{2}(s-{y}_{0}){\}}_{s}^{\prime}|ds}$$14(14)Combining (13) and (14), we obtain$$|{i}_{\lambda}(x,y,t)|\underset{\_}{<}{\u03f5}^{p}\underset{{y}_{0}-\delta}{\overset{\mathcal{y}0+\delta}{\int}}{L}_{\lambda}(t-x,s-y)|\{{\mu}_{2}(|s-{y}_{0}|){\}}_{t}^{\prime}|ds+2{\mu}_{2}(|y-{y}_{0}|){L}_{\lambda}(t-x,0).$$

Similar calculations for the integral ${j}_{\lambda}(x,y,s)$yield$$|{j}_{\lambda}(x,y,s)|\underset{\_}{<}{\u03f5}^{p}\underset{{x}_{0}-\delta}{\overset{x\mathrm{o}+\delta}{\int}}{L}_{\lambda}(t-x,s-y)\left|\{{\mu}_{1}(|t-{x}_{0}|){\}}_{t}^{\prime}|dt+2{\mu}_{1}(|x-{x}_{0}\right|){L}_{\lambda}(0,s-y).$$

Thus, we have$${I}_{31}(x,y,\lambda )\underset{\_}{<}{\u03f5}^{p}{2}^{p}\underset{{x}_{0}-\delta}{\overset{{x}_{0}+\delta}{\int}}\left[\underset{{y}_{0}-\delta}{\overset{{\mathcal{y}}_{0}+\delta}{\int}}{L}_{\lambda}(t-x,s-y)|\{{\mu}_{2}(|s-{y}_{0}|){\}}_{t}^{\prime}|ds+2{\mu}_{2}(|y-{y}_{0}|){L}_{\lambda}(t-x,0)\right]dt+{\u03f5}^{p}{2}^{p}\underset{{y}_{0}-\delta}{\overset{{y}_{0}+\delta}{\int}}\left[\underset{{x}_{0}-\delta}{\overset{{x}_{0}+\delta}{\int}}{L}_{\lambda}(t-x,s-y)|\{{\mu}_{1}(|t-{x}_{0}|){\}}_{t}^{\prime}|dt+2{\mu}_{1}(|x-{x}_{0}|){L}_{\lambda}(0,s-y)\right]ds.$$

Since *epsilon* is arbitrary and *L*_{λ} is integrable with respect to
eachvariable, the desired result follows from hypotheses (8)and (7), i.e.,$(x,y,\lambda )\to ({x}_{0},{y}_{0},{\lambda}_{0})$

Note that the same
conclusion is obtained for the case $0<y-{y}_{0}<{\displaystyle \frac{\delta}{2}}$ Thus
the proof is completed.

## Comments (0)

General note:By using the comment function on degruyter.com you agree to our Privacy Statement. A respectful treatment of one another is important to us. Therefore we would like to draw your attention to our House Rules.